Question

A variable plane keeping at a constant distance $P$ from the origin $O$ cuts the intercept on the axis at the points $A,B,C$. Show that the locus of centroid of tetrahedron $OABC$ is $x^{-2}+y^{-2}+z^{-2}=16P^{-2}$

Answer 1

$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}+=\mathrm{1....}\left(1\right)$
Since, the plane is at a distance of $p$ from origin
$p=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}$
$\frac{1}{p^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}......\left(2\right)$
Plane cuts axes at $A(a,0,0),B(0,b,0)$ and $C(0,0,c)$
Let $(x,y,z)$ be the co-ordinates of centroid of tetrahedron then
$x=\frac{a}{4},y=\frac{b}{4},z=\frac{c}{4}$,
$a=4x,b=4y,c=4z$
Putting these value in equation (2)
$\frac{1}{p^2}=\frac{1}{16x^2}+\frac{1}{16y^2}+\frac{1}{16z^2}$
$\frac{16}{p^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}$