Q67 / 120 · VITEEE 2016 · Mathematics
A variable plane remains at constant distance $\mathrm{p}$, from the origin.If it meets coordinate axes at points $A, B, C$ then the locus of the centroid of $\Delta$ $\mathrm{ABC}$ is
- Correct answerA$\mathrm{x}^{-2}+\mathrm{y}^{-2}+\mathrm{z}^{-2}=9 \mathrm{p}^{-2}$
- B$x^{-3}+y^{-3}+z^{-3}=9 p^{-3}$
- C$x^{2}+y^{2}+z^{2}=9 p^{2}$
- D$x^{3}+y^{3}+z^{3}=9 p^{3}$
Solution
Let $\mathrm{A} \equiv(\mathrm{a}, 0,0), \mathrm{B} \equiv(0, \mathrm{~b}, 0), \mathrm{C} \equiv(0,0, \mathrm{c})$, then equation of the plane is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
Its distance from the origin, $\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}=\frac{1}{p^{2}} \ldots$ (i)
If $(x, y, z)$ be centroid of $\Delta A B C$, then
$$
\mathrm{x}=\frac{\mathrm{a}}{3}, \mathrm{y}=\frac{\mathrm{b}}{3}, \mathrm{z}=\frac{\mathrm{c}}{3}...(ii)
$$
Eliminating a,b,c from (i) and (ii) required locus is
$x^{-2}+y^{-2}+z^{-2}=9 p^{-2}$
Hence, the correct answer is option: 'A'.