The perpendicular distance of the point $P (6, 7, 8)$ from $X Y -$ plane is

8

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7

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6

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5

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Solution

The correct option is A $8$
The perpendicular distance of a point $(x, y, z)$ from $X Y -$ plane is $| z |$ .
So, the perpendicular distance of the point $P (6, 7, 8)$ from $X Y -$ plane is $8$ units.

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