$P$ is a point such that the sum of the squares of its distances from the planes $x + y + z = 0$ , $x + y - 2 z = 0, x - y = 0$ is $5$ , then the locus of $P$ is

x^{2} + y^{2} + z^{2} = 10

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x^{2} + y^{2} + z^{2} = 25

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x^{2} + y^{2} + z^{2} = 5

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x^{2} + y^{2} + z^{2} = 50

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Solution

The correct option is C $x^{2} + y^{2} + z^{2} = 5$

Let $P (x, y, z)$ be the point

Now, from perpendicular distance formula ${(\frac{x + y - z}{\sqrt{3}})}^{2} + {(\frac{x + y - 2 z}{\sqrt{6}})}^{2} + {(\frac{x - y}{\sqrt{2}})}^{2} = 5$

Therefore, $x^{2} + y^{2} + z^{2} = 5$

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