The distance between the point $P (x, y, z)$ and plane $x z$ is :

x

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y

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z

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x y

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Solution

The correct option is A $y$
The coordinate of any point in xz plane is $Q (x, 0, y)$
Thus, the distance between point P and Q is $= \sqrt{(x - x)^{2} + (y - 0)^{2} + (z - z)^{2}} = y$

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