Let distances of a point $P\left(x,y,z\right)$ from the planes $x+y+z=0, x-z=0$ and $x-2y+z=0$ are $\frac{x+y+z}{\sqrt{3}}, \frac{x-z}{\sqrt{2}}$ and $\frac{x-2y+z}{\sqrt{6}}$ respectively, then the sum of the squares of distances, is as

$\Rightarrow {\left(\frac{x+y+z}{\sqrt{3}}\right)}^2+{\left(\frac{x-z}{\sqrt{2}}\right)}^2+{\left(\frac{x-2y+z}{\sqrt{6}}\right)}^2=p^2$

$\Rightarrow 2{\left(x+y+z\right)}^2+3{\left(x-z\right)}^2+{\left(x-2y+z\right)}^2=6p^2$

$\Rightarrow 2x^2+2y^2+2z^2+4xy+4yz+4zx+3x^2+3z^2-6xz+x^2+4y^2+z^2-4xy-4yz+2xz=6p^2$

$\Rightarrow 6x^2+6y+6z^2=6p^2$

$\Rightarrow x^2+y^2+z^2=p^2$

Hence, option (d) is correct.