Thank you for reporting, we will resolve it shortly
Q.
The locus of a point, such that the sum of the squares of its distances from the planes $x+y+z=0, x-z=0$ and $x -2 y + z =0$ is $9$, is
Three Dimensional Geometry
Solution:
Let the variable point be $(\alpha, \beta, \gamma)$ then according to question
$\left(\frac{|\alpha+\beta+\gamma|}{\sqrt{3}}\right)^{2}+\left(\frac{|\alpha-\gamma|}{\sqrt{2}}\right)^{2}+\left(\frac{|\alpha-2 \beta+\gamma|}{\sqrt{6}}\right)^{2}=9$
$\Rightarrow \alpha^{2}+\beta^{2}+\gamma^{2}=9$.
So, the locus of the point is $x^{2}+y^{2}+z^{2}=9$