ML Aggarwal Class 11 Maths Solutions Chapter 12 Circles

Chapter 12: Introduction to Three Dimensional Geometry

12.1 Cartesian Co-ordinate System

In two-dimensional geometry, any point in a plane is represented using an ordered pair of numbers (x, y) with reference to two perpendicular axes called the x-axis and y-axis. The extension of this concept to three dimensions involves using three mutually perpendicular axes.

Co-ordinate Axes

Three mutually perpendicular directed lines intersect at a fixed point O. When these lines are coordinated using O as the origin and the same unit of length on each line:

• X'OX is the x-axis
• Y'OY is the y-axis
• Z'OZ is the z-axis

These three axes taken together are called the coordinate axes or simply axes. Since they are mutually at right angles to each other, they are also called rectangular axes. The fixed point O is called the origin.

Co-ordinate Planes

Three coordinate planes are formed by pairs of axes:

• The XY-plane contains the x-axis and y-axis
• The YZ-plane contains the y-axis and z-axis
• The ZX-plane contains the z-axis and x-axis

Co-ordinates of a Point

Consider any point P in space. If planes are drawn through P parallel to the coordinate planes, these planes meet the axes at points A, B, and C respectively. If the directed distances OA, OB, and OC are x, y, and z respectively, then the ordered triplet (x, y, z) represents the Cartesian coordinates of P, written as P(x, y, z).

Given any point P in space, there is a unique ordered triplet (x, y, z) of real numbers. Conversely, any ordered triplet (x, y, z) of real numbers corresponds to exactly one unique point in space. This means there exists a one-to-one correspondence between points in space and ordered triplets of real numbers.

Key Remarks:

• The origin has coordinates O(0, 0, 0)
• The x-coordinate represents the directed distance of P from the YZ-plane
• The y-coordinate represents the directed distance of P from the ZX-plane
• The z-coordinate represents the directed distance of P from the XY-plane
• Any point on the XY-plane has the form (x, y, 0), and the equation of the XY-plane is z = 0
• Any point on the YZ-plane has the form (0, y, z), and the equation of the YZ-plane is x = 0
• Any point on the ZX-plane has the form (x, 0, z), and the equation of the ZX-plane is y = 0
• Any point on the x-axis has the form (x, 0, 0); the equations of the x-axis are y = 0 and z = 0
• Any point on the y-axis has the form (0, y, 0); the equations of the y-axis are x = 0 and z = 0
• Any point on the z-axis has the form (0, 0, z); the equations of the z-axis are x = 0 and y = 0

Octants

The three coordinate planes divide space into eight regions called octants. The signs of a point's coordinates determine which octant it lies in. The table below shows the sign pattern for coordinates in each octant:

Question 1. In fig. 12.2, if the point P is (3, 2, 4), find the co-ordinates of the points M, N and L.
Answer: For point M, the distance along the z-direction is zero. So M has coordinates (3, 2, 0). For point N, the distance along the x-direction is zero. Therefore, N is located at (0, 2, 4). Similarly, point L, where the y-distance is zero, has coordinates (3, 0, 4).
In simple words: To find each point, set one coordinate to zero and keep the other two the same as P.

Exam Tip: When a question refers to a figure with projection points, identify which axis direction corresponds to zero for each point, then keep the other two coordinates unchanged.

Question 2. If L, M and N be the feet of perpendiculars drawn from a point P(3, 4, 5) on the x, y and z-axes respectively, then find the co-ordinates of L, M and N.
Answer: The foot of the perpendicular from P(3, 4, 5) to the x-axis is the point on the x-axis closest to P. On the x-axis, both y and z coordinates are zero, so L is (3, 0, 0). Similarly, M on the y-axis has coordinates (0, 4, 0), and N on the z-axis has coordinates (0, 0, 5).
In simple words: On each axis, only one coordinate is non-zero. The foot of perpendicular has the same value for that one coordinate and zeros for the other two.

Exam Tip: Remember that on each axis, two coordinates must be zero - keep only the coordinate that matches the axis direction.

Question 3. In fig. 12.2, if the point P is (4, 2, 3), find the co-ordinates of the reflection (mirror image) of the point P in the XY-plane.
Answer: The XY-plane is where z = 0. When a point is reflected in this plane, the image lies as far below the plane as the original point lies above it. Since P is at height 3 above the XY-plane, its reflection is at height 3 below, giving a z-coordinate of -3. The x and y coordinates remain the same. Therefore, the reflection of P(4, 2, 3) in the XY-plane is (4, 2, -3).
In simple words: For reflection in the XY-plane, keep x and y the same but make the z-coordinate negative.

Exam Tip: When reflecting in a coordinate plane, the coordinates parallel to that plane stay the same, while the perpendicular coordinate changes sign.

Question 4. Find the octants in which the points (-4, -2, 5) and (-3, 2, -1) lie.
Answer: For the point (-4, -2, 5), the x-coordinate is negative, y-coordinate is negative, and z-coordinate is positive. This sign pattern (-, -, +) matches the third octant. For the point (-3, 2, -1), the x-coordinate is negative, y-coordinate is positive, and z-coordinate is negative. This sign pattern (-, +, -) matches the sixth octant.
In simple words: Check the signs of all three coordinates and match them to the octant table to find which octant contains each point.

Exam Tip: Memorize or write down the sign patterns for all eight octants - this is crucial for quickly identifying which octant a point lies in.

Question 5. What are the co-ordinates of the vertices of a cube whose edge is 2 units, one of whose vertices coincides with origin and the three edges passing through the origin coincides with the positive direction of axes through the origin.
Answer: The cube has one vertex at the origin O and extends 2 units along each positive axis. The eight vertices of the cube are positioned as follows: O(0, 0, 0) is at the origin; A(2, 0, 0) extends 2 units along the positive x-axis; B(0, 2, 0) extends 2 units along the positive y-axis; C(0, 0, 2) extends 2 units along the positive z-axis. The remaining vertices are D(2, 2, 0) formed by combining the x and y extensions; E(0, 2, 2) combining y and z extensions; F(2, 0, 2) combining x and z extensions; and G(2, 2, 2) at the far corner, combining all three extensions.
In simple words: Each vertex has coordinates that are either 0 or 2, depending on whether that edge extends along that axis.

Exam Tip: Draw a simple 3D cube sketch and label each vertex systematically - start with O at origin, mark single-axis extensions, then double-axis extensions, and finally the far corner.

Exercise 12.1 - Very Short Answer Type Questions

Question 1. A point is on the x-axis. What are its y-coordinate and z-coordinate?
Answer: Both the y-coordinate and z-coordinate are zero. Any point on the x-axis has the form (x, 0, 0).
In simple words: Points on the x-axis have zero values for both the y and z directions.

Exam Tip: Remember that an axis is the intersection of two coordinate planes set to zero - only the direction of that axis varies.

Question 2. If a point lies on y-axis, then what are its x-coordinate and z-coordinate?
Answer: Both the x-coordinate and z-coordinate are zero. Any point lying on the y-axis has the form (0, y, 0).
In simple words: On the y-axis, only the y-value changes; both x and z are always zero.

Exam Tip: Apply the same logic consistently - each axis represents movement in one direction only, with the other two directions at zero.

Question 3. If a point lies in the YZ-plane, then what can you say about its x-coordinate?
Answer: The x-coordinate of any point in the YZ-plane is zero. The YZ-plane is the plane where x = 0, so all points in this plane can be written in the form (0, y, z).
In simple words: In the YZ-plane, the x-direction is always zero because this plane is perpendicular to the x-axis.

Exam Tip: Each coordinate plane is defined by setting one coordinate to zero - the YZ-plane has x = 0, the ZX-plane has y = 0, and the XY-plane has z = 0.

Question 4. A point is in the ZX-plane. What can you say about its y-coordinate?
Answer: The y-coordinate of any point in the ZX-plane is zero. Points in the ZX-plane have the form (x, 0, z).
In simple words: The ZX-plane is perpendicular to the y-axis, so all points in it have y = 0.

Exam Tip: Identify which axis is perpendicular to the plane in question - that coordinate will always be zero.

Question 5. Write the coordinates of the foot of perpendicular from the point (3, -2, 5) on (i) the x-axis (ii) the y-axis (iii) the z-axis.
Answer:
(i) The foot of perpendicular on the x-axis is (3, 0, 0) - the x-coordinate stays the same while y and z become zero.
(ii) The foot of perpendicular on the y-axis is (0, -2, 0) - the y-coordinate stays the same while x and z become zero.
(iii) The foot of perpendicular on the z-axis is (0, 0, 5) - the z-coordinate stays the same while x and y become zero.
In simple words: For each axis, keep only the coordinate matching that axis and set the other two to zero.

Exam Tip: The foot of perpendicular on an axis retains only the coordinate for that axis direction.

Question 6. Write the coordinates of the foot of perpendicular from the point (3, -4, 5) on the (i) XY-plane (ii) YZ-plane (iii) ZX-plane.
Answer:
(i) The foot of perpendicular on the XY-plane is (3, -4, 0) - the z-coordinate becomes zero since the XY-plane is at z = 0.
(ii) The foot of perpendicular on the YZ-plane is (0, -4, 5) - the x-coordinate becomes zero since the YZ-plane is at x = 0.
(iii) The foot of perpendicular on the ZX-plane is (3, 0, 5) - the y-coordinate becomes zero since the ZX-plane is at y = 0.
In simple words: For each plane, set the perpendicular coordinate to zero and keep the other two unchanged.

Exam Tip: Each coordinate plane is perpendicular to one axis - that axis's coordinate becomes zero in the foot of perpendicular.

Question 7. Write the co-ordinates of the image of the point (-3, 2, 7) in the (i) XY-plane (ii) YZ-plane (iii) ZX-plane.
Answer:
(i) The image in the XY-plane is (-3, 2, -7) - the z-coordinate changes sign.
(ii) The image in the YZ-plane is (3, 2, 7) - the x-coordinate changes sign.
(iii) The image in the ZX-plane is (-3, -2, 7) - the y-coordinate changes sign.
In simple words: When reflecting in a plane, the coordinate perpendicular to that plane reverses its sign, while the other two stay the same.

Exam Tip: For reflection in a coordinate plane, only the coordinate perpendicular to the plane changes sign - identify which coordinate is perpendicular and negate it.

Question 8. Find the (mirror) image of the given point in the specified plane: (i) (-3, 4, 7) in the YZ-plane (ii) (-7, 2, -1) in the ZX-plane (iii) (5, 4, -3) in the XY-plane (iv) (-4, 0, 1) in the ZX-plane (v) (-2, 0, 0) in the XY-plane.
Answer:
(i) Image of (-3, 4, 7) in the YZ-plane: (3, 4, 7) - the x-coordinate changes sign.
(ii) Image of (-7, 2, -1) in the ZX-plane: (-7, -2, -1) - the y-coordinate changes sign.
(iii) Image of (5, 4, -3) in the XY-plane: (5, 4, 3) - the z-coordinate changes sign.
(iv) Image of (-4, 0, 1) in the ZX-plane: (-4, 0, 1) - the y-coordinate is already zero, so it stays the same.
(v) Image of (-2, 0, 0) in the XY-plane: (-2, 0, 0) - the z-coordinate is already zero, so the point remains unchanged.
In simple words: Identify the perpendicular coordinate to the plane. If it is non-zero, change its sign. If it is already zero, the point does not move.

Exam Tip: When a point already lies on the plane, its reflection is itself - this happens when the perpendicular coordinate is zero.

Question 9. Name the octant in which the following points lie: (i) (1, 3, 5) (ii) (3, -2, 4) (iii) (-2, -3, 5) (iv) (-1, -2, -5) (v) (-3, 1, 2) (vi) (2, -1, -5) (vii) (5, 2, -3) (viii) (-1, 3, -2).
Answer:
(i) (1, 3, 5): x positive, y positive, z positive - Octant I
(ii) (3, -2, 4): x positive, y negative, z positive - Octant IV
(iii) (-2, -3, 5): x negative, y negative, z positive - Octant III
(iv) (-1, -2, -5): x negative, y negative, z negative - Octant VII
(v) (-3, 1, 2): x negative, y positive, z positive - Octant II
(vi) (2, -1, -5): x positive, y negative, z negative - Octant VIII
(vii) (5, 2, -3): x positive, y positive, z negative - Octant V
(viii) (-1, 3, -2): x negative, y positive, z negative - Octant VI
In simple words: Write down the sign of each coordinate. Then use the octant table to find which octant matches that sign pattern.

Exam Tip: Refer to the octant table systematically and check all three coordinate signs before identifying the octant.

Question 10. In fig. 12.2, if the point P is (a, b, c), write the co-ordinates of the points M, N, L, A, B and C.
Answer: Based on the figure, M is on the XY-plane (directly below P), so M has coordinates (a, b, 0). N is on the x-axis where a perpendicular from M meets it, so N is (a, 0, 0). L is on the z-axis reached by a perpendicular from P, giving L as (0, 0, c). Point A is on the x-axis at (a, 0, 0). Point B is on the y-axis at (0, b, 0). Point C is on the z-axis at (0, 0, c).
In simple words: Project the point P onto each plane and axis - each projection gives a point with some coordinates replaced by zero.

Exam Tip: Use systematic projection: set one or two coordinates to zero depending on whether you want a projection on a plane or an axis.

Question 11. Fill in the blanks: (i) The x-axis and y-axis taken together determine a plane known as .............. (ii) The co-ordinates of a point in the XY-plane are of the form .............. (iii) Co-ordinate planes divide the space into .............. octants.
Answer:
(i) The x-axis and y-axis together determine the XY-plane.
(ii) The coordinates of any point in the XY-plane are of the form (x, y, 0).
(iii) The coordinate planes divide space into eight octants.
In simple words: (i) Two axes form a plane. (ii) A plane has one coordinate always zero. (iii) Three perpendicular planes create eight regions.

Exam Tip: These are fundamental definitions - memorize them as they form the foundation for 3D coordinate geometry.

12.2 Distance Formula

To calculate the distance between two points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) in three-dimensional space, we construct a rectangular box by drawing planes parallel to the coordinate planes through P and Q. The diagonal PQ of this box can be found using the Pythagorean theorem applied twice.

Consider the box formed: The distance along the x-direction is |x₂ - x₁|, along the y-direction is |y₂ - y₁|, and along the z-direction is |z₂ - z₁|. Using the three-dimensional Pythagorean theorem:

Distance Formula: PQ = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²]

As a special case, the distance from a point P(x₁, y₁, z₁) to the origin O(0, 0, 0) is:

Distance from Origin: OP = √(x₁² + y₁² + z₁²)

Locus of a Point: The locus of a point in space is the set of all positions that a moving point can occupy when it satisfies a given geometrical condition or set of conditions. This forms a surface in three-dimensional space.

Question 1. Find the distance between the points P(-3, 7, 2) and Q(2, 4, -1).
Answer: Using the distance formula, PQ = √[(2 - (-3))² + (4 - 7)² + (-1 - 2)²] = √[(5)² + (-3)² + (-3)²] = √[25 + 9 + 9] = √43 units.
In simple words: Find the change in each coordinate, square each change, add them together, and take the square root of the sum.

Exam Tip: Write out the differences for each coordinate first, then square and add them - this reduces arithmetic errors.

Question 2. If the distance between the points (a, 2, 1) and (1, -1, 1) is 5 units, find a.
Answer: Using the distance formula, √[(1 - a)² + (-1 - 2)² + (1 - 1)²] = 5. Simplifying, √[(1 - a)² + 9 + 0] = 5. Squaring both sides gives (1 - a)² + 9 = 25. Therefore (1 - a)² = 16, which means 1 - a = ±4. Solving these: if 1 - a = 4, then a = -3; if 1 - a = -4, then a = 5. Thus a = 5 or a = -3.
In simple words: Set up the distance formula, square both sides to remove the square root, then solve the resulting equation for the unknown.

Exam Tip: When you get ±4 after taking the square root, remember to check both values - both are valid solutions unless context eliminates one.

Question 3. Find the length of perpendicular drawn from the point P(3, 4, 5) on y-axis.
Answer: The foot of the perpendicular from P(3, 4, 5) to the y-axis lies on the y-axis. Any point on the y-axis has the form (0, y, 0). Since the perpendicular foot has the same y-coordinate as P, it is at M(0, 4, 0). The length of this perpendicular is the distance PM = √[(3 - 0)² + (4 - 4)² + (5 - 0)²] = √[9 + 0 + 25] = √34 units.
In simple words: Find the foot of perpendicular on the axis by keeping the matching coordinate and setting the others to zero. Then calculate the distance from P to this foot.

Exam Tip: For perpendicular distance to an axis, identify the foot by keeping one coordinate and zeroing the others, then use the distance formula.

Question 4. If a parallelopiped is formed by planes drawn through the points P(2, 3, 5) and Q(5, 9, 7) parallel to the co-ordinate planes, then find the lengths of edges of the parallelopiped and the length of a diagonal.
Answer: When planes parallel to the coordinate planes are drawn through P and Q, they form a rectangular box (cuboid) with P and Q at opposite corners. The edge parallel to the x-axis has length |5 - 2| = 3 units. The edge parallel to the y-axis has length |9 - 3| = 6 units. The edge parallel to the z-axis has length |7 - 5| = 2 units. The main diagonal connecting P and Q has length PQ = √[(5 - 2)² + (9 - 3)² + (7 - 5)²] = √[9 + 36 + 4] = √49 = 7 units.
In simple words: The edge lengths are the differences in each coordinate. The diagonal is found using the distance formula applied to the two opposite corners.

Exam Tip: The coordinate differences give you the box dimensions - visualizing this rectangular box helps understand why the diagonal uses the distance formula.

Question 5. By using distance formula, show that the points P(-2, 3, 5), Q(1, 2, 3) and R(7, 0, -1) are collinear.
Answer: Calculate PQ = √[(1 - (-2))² + (2 - 3)² + (3 - 5)²] = √[9 + 1 + 4] = √14 units. Calculate QR = √[(7 - 1)² + (0 - 2)² + (-1 - 3)²] = √[36 + 4 + 16] = √56 = 2√14 units. Calculate PR = √[(7 - (-2))² + (0 - 3)² + (-1 - 5)²] = √[81 + 9 + 36] = √126 = 3√14 units. Since PR = PQ + QR (as 3√14 = √14 + 2√14), the point Q lies on the line segment PR. Therefore, the three points are collinear.
In simple words: Calculate all three pairwise distances. If the longest distance equals the sum of the other two, the three points lie on the same straight line.

Exam Tip: For collinearity, the sum of two shorter distances must exactly equal the longest distance - this is the key check.

Question 6. Show that the points A(0, 7, 10), B(-1, 6, 6) and C(-4, 9, 6) are the vertices of an isosceles right angled triangle.
Answer: First, calculate the three side lengths. AB = √[(-1 - 0)² + (6 - 7)² + (6 - 10)²] = √[1 + 1 + 16] = √18 units. BC = √[(-4 - (-1))² + (9 - 6)² + (6 - 6)²] = √[9 + 9 + 0] = √18 units. CA = √[(-4 - 0)² + (9 - 7)² + (6 - 10)²] = √[16 + 4 + 16] = √36 = 6 units. Since AB = BC = √18, the triangle is isosceles. Check if it is right-angled: AB² + BC² = 18 + 18 = 36 = CA², which confirms the Pythagorean theorem. Therefore, triangle ABC is isosceles and right-angled at B.
In simple words: Find all three sides. Check if any two are equal (isosceles). Then check if the Pythagorean theorem holds to confirm the right angle.

Exam Tip: Always verify both conditions for an isosceles right triangle - equal sides AND the Pythagorean relation - rather than assuming one implies the other.

Question 7. Are the points A(3, 6, 9), B(10, 20, 30) and C(25, -41, 5), the vertices of a right angled triangle?
Answer: Calculate the three distances. AB = √[(10 - 3)² + (20 - 6)² + (30 - 9)²] = √[49 + 196 + 441] = √686 units. BC = √[(25 - 10)² + (-41 - 20)² + (5 - 30)²] = √[225 + 3721 + 625] = √4571 units. CA = √[(3 - 25)² + (6 - (-41))² + (9 - 5)²] = √[484 + 2209 + 16] = √2709 units. Since BC is the longest side, check if AB² + CA² = BC². We find AB² + CA² = 686 + 2709 = 3395, but BC² = 4571. Since 3395 ≠ 4571, the Pythagorean theorem does not hold. Therefore, the triangle is not a right-angled triangle.
In simple words: Calculate all three sides, identify the longest one, then check if the Pythagorean relation holds for these three measurements.

Exam Tip: When checking for a right angle, always ensure the Pythagorean relation uses the longest side as the hypotenuse - if the relation fails, there is no right angle.

Chapter Test

Question 1. By using distance formula, prove that the points A(3, -5, 1), B(-1, 0, 8) and C(7, -10, -6) are collinear.
Answer: Calculate AB = √[(-1 - 3)² + (0 - (-5))² + (8 - 1)²] = √[16 + 25 + 49] = √90 units. Calculate BC = √[(7 - (-1))² + (-10 - 0)² + (-6 - 8)²] = √[64 + 100 + 196] = √360 = 2√90 units. Calculate AC = √[(7 - 3)² + (-10 - (-5))² + (-6 - 1)²] = √[16 + 25 + 49] = √90 units. Wait, let me recalculate AC: √[16 + 25 + 49] = √90. Actually, this seems incorrect. Let me recalculate more carefully. AC = √[(7-3)² + (-10-(-5))² + (-6-1)²] = √[16 + 25 + 49] = √90. This gives AB = AC = √90 and BC = 2√90. Since AC + AB = √90 + √90 = 2√90 = BC, the three points are collinear with B not between A and C but rather C beyond the line on the opposite side. Actually, if AC + AB = BC, then B is between A and C. Let me verify: if AB + BC = AC, then B is between A and C. Here AB + BC would be √90 + 2√90 ≠ AC. Let me reconsider: if AB + AC = BC is false, let's check if the distances satisfy collinearity differently. Actually AB = √90, BC = 2√90, so AB + BC ≠ AC is possible only if they're collinear with a different order. The key is that these should satisfy one ordering. Since BC = 2√90 and AB + AC would need to be calculated... actually I made an error. Let me recalculate AC correctly: AC = √[(7-3)² + (-10+5)² + (-6-1)²] = √[16 + 25 + 49] = √90. Hmm, this equals AB. This suggests reflection rather than collinearity as I set it up. Let me recalculate from scratch with the original points: A(3, -5, 1), B(-1, 0, 8), C(7, -10, -6). AB = √[(-1-3)² + (0+5)² + (8-1)²] = √[16 + 25 + 49] = √90. BC = √[(7+1)² + (-10-0)² + (-6-8)²] = √[64 + 100 + 196] = √360 = 6√10. AC = √[(7-3)² + (-10+5)² + (-6-1)²] = √[16 + 25 + 49] = √90. We have AB = AC = √90 and BC = 6√10 ≈ 18.97, while AB + AC = 2√90 ≈ 18.97. Let me check: 2√90 = √360 = 6√10. Yes! So AB + AC = BC, meaning points are collinear.
In simple words: Calculate the three distances. If the sum of the two smaller distances equals the largest distance, then the three points lie on the same straight line.

Exam Tip: When proving collinearity, calculate all three pairwise distances and verify that the sum of two equals the third - this is the definitive test.

Question 2. Find the points on z-axis which are at a distance √21 from the point A(1, 2, 3).
Answer: Any point on the z-axis has the form (0, 0, z). The distance from this point to A(1, 2, 3) is √[(0-1)² + (0-2)² + (z-3)²] = √[1 + 4 + (z-3)²] = √[5 + (z-3)²]. Setting this equal to √21: √[5 + (z-3)²] = √21, which gives 5 + (z-3)² = 21, so (z-3)² = 16. Therefore z - 3 = ±4, giving z = 7 or z = -1. The required points are (0, 0, 7) and (0, 0, -1).
In simple words: Write a general point on the z-axis, apply the distance formula, set it equal to the given distance, and solve for the z-coordinate.

Exam Tip: When finding points on an axis at a specific distance from another point, parameterize the axis point and use the distance formula as an equation to solve.

Question 3. Find the co-ordinates of the point which is equidistant from the four points with coordinates O(0, 0, 0), A(-3, 0, 0), B(0, 2, 0) and C(0, 0, -5). Find also the distance of this point from each of the given points.
Answer: Let the required point be P(x, y, z). Since P is equidistant from all four points, we have PO = PA = PB = PC. From PO = PA: x² + y² + z² = (x+3)² + y² + z², which simplifies to x² = x² + 6x + 9, giving 6x = -9, so x = -3/2. From PO = PB: x² + y² + z² = x² + (y-2)² + z², which simplifies to y² = y² - 4y + 4, giving 4y = 4, so y = 1. From PO = PC: x² + y² + z² = x² + y² + (z+5)², which simplifies to z² = z² + 10z + 25, giving 10z = -25, so z = -5/2. The point is P(-3/2, 1, -5/2). The distance from P to O is √[(-3/2)² + 1² + (-5/2)²] = √[9/4 + 1 + 25/4] = √[38/4] = √(19/2) units. By symmetry of the construction, this distance is the same from P to each of the other three points.
In simple words: Set the distances from the unknown point P to each of the four given points equal to each other. Solve the three resulting equations to find the coordinates of P, then calculate the distance.

Exam Tip: Use pairwise equality of distances (d(P,O) = d(P,A), d(P,O) = d(P,B), d(P,O) = d(P,C)) to generate three independent equations - each equation eliminates one variable.