Question

Find the locus of the point, the sum of the squares of whose distances from the planes $x+y+z=0$, $x-y=0$, $x+y-2z=0$ is $7$.

Answer 1

$P_1:x+y+z=0,P_2:x-y=0,P_3:x+y-2z=0$

Let $(h,k,l)$ be the point

$\therefore$ distance for $P_1$,

$d_1=\frac{h+k+l}{\sqrt{3}}$

$\therefore$ Distance from $P_2$,$d_2=\frac{h-k}{\sqrt{2}}$

$\therefore$ from $P_3$, $d_3=\frac{h+k-2l}{\sqrt{2^2+1^2+1^2}}=\frac{h+k-l}{\sqrt{6}}$

$d_1^2+d_2^2+d_3^2=7\frac{{(h+k+l)}^2}{3}+\frac{{(h-k)}^2}{2}+\frac{{(h+k-2l)}^2}{6}=72{(h+k+l)}^2+3{(h-k)}^2+{(h+k-2l)}^2=7\Rightarrow 2(h^2+k^2+l^2+2hk+2hl+2kl)+3(h^2+k^2-2hk)+(h^2+k^2+4l^2+2hk-4kl-4hl)=7\Rightarrow {6h}^2+6k^2+6l^2+0hk+0hl+0kl=7\Rightarrow h^2+k^2+l^2=7$

Locus of required point is,

$x^2+y^2+z^2=7$