The locus of a point, such that the sum of the squares of its distance from the planes $x + y + z = 0, x - z = 0$ and $x - 2 y + z = 0$ is $9$ , is

x^{2} + y^{2} + z^{2} = 3

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x^{2} + y^{2} + z^{2} = 6

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x^{2} + y^{2} + z^{2} = 9

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x^{2} + y^{2} + z^{2} = 12

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Solution

The correct option is D $x^{2} + y^{2} + z^{2} = 9$
Let $(a, b, c)$ be the point satisfy the given condition
$\Rightarrow \frac{(a + b + c)^{2}}{3} + \frac{(a - c)^{2}}{2} + \frac{(a - 2 b + c)^{2}}{6} = 9$
$\Rightarrow a^{2} + b^{2} + c^{2} = 9$
$\Rightarrow x^{2} + y^{2} + z^{2} = 9$

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