AP EAMCET · Maths · Three Dimensional Geometry
The distance from a point \((1,1,1)\) to a variable plane \(\pi\) is 12 units and the points of intersections of the plane \(\pi\) and \(\mathrm{X}, \mathrm{Y}, \mathrm{Z}\)-axes are \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) respectively. If the point of intersection of the planes through the points \(\mathrm{A}, \mathrm{B}, \mathrm{C}\) and parallel to the coordinate planes is P , then the equation of the locus of P is
- A \(\left(\frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}\right)=143\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
- B \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=144\)
- C \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)^2=144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
- D \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)^2=144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)^2\)
Answer & Solution
Correct Answer
(C) \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-1\right)^2=144\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
Step-by-step Solution
Detailed explanation
Let the equation of the required plane be \(\mathrm{T}_1: \frac{x}{a}+\frac{y}{b}+\frac{z}{c}-1=0 \Rightarrow\left|\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=12\) \(P \equiv(a, b, c)\)…
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