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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 $x - 2y + z = 0$ is 9, is
Three Dimensional Geometry
Solution:
Let the variable point be $\left( \alpha, \beta, \gamma\right)$ then according to question
$\left(\frac{\left|\alpha+\beta+\gamma\right|}{\sqrt{3}}\right)^{2} + \left(\frac{\left|\alpha -\gamma \right|}{\sqrt{2}}\right)^{2} + \left(\frac{\left|\alpha -2\beta +\gamma \right|}{\sqrt{6}}\right)^{2}= 9$
$\Rightarrow \alpha^{2} +\beta^{2} +\gamma^{2} = 9$.
So, the locus of the point is $x^{2}+y^{2} + z^{2} = 9$