Question

The distance from a point \( (1,1,1) \) to a variable plane \(\pi\) is 12 units and the points of intersections of the plane with X, Y, Z-axes are \( A, B, C \) respectively. If the point of intersection of the planes through the points \( A, B, C \) and parallel to the coordinate planes is \( P \), then the equation of the locus of \( P \) is: 

• A. \( \left( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \right) = 143 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right) \)
• B. \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 144 \)
• C. \( \left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)^2 = 144 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right) \)
• D. \( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 144 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right)^2 \)

Hint:

For problems involving locus of points derived from intersection of coordinate planes, use intercept form equations and apply distance formulas carefully.

Answer 1

The Correct Option is C

Step 1: Equation of the Plane 
The general equation of a plane passing through a given point \( (a, b, c) \) and having intercepts \( A, B, C \) on the X, Y, and Z axes respectively is: \[ \frac{x}{A} + \frac{y}{B} + \frac{z}{C} = 1. \] Given that the perpendicular distance from the point \( (1,1,1) \) to this plane is 12, we use the formula for the distance from a point to a plane: \[ \frac{|1/A + 1/B + 1/C - 1|}{\sqrt{(1/A)^2 + (1/B)^2 + (1/C)^2}} = 12. \] Squaring both sides and simplifying: \[ \left( \frac{1}{A} + \frac{1}{B} + \frac{1}{C} \right)^2 = 144 \left( \frac{1}{A^2} + \frac{1}{B^2} + \frac{1}{C^2} \right). \] Since \( P \) is the intersection of planes parallel to the coordinate planes passing through \( A, B, C \), its coordinates satisfy the same relation. 
Final Answer: \( \boxed{\left( \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \right)^2 = 144 \left( \frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2} \right)} \)