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Question

Find the distance of point (1,2,3) from the plane xy+z=5, measured parallel to the line whose direction cosines are proportional to 2,3,6.

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Solution

Here, we are not required to find the perpendicular distance of the point (1,2,3) from the plane, but the distance measured parallel to line whose D.R.'s are 2,3,6.
Equation of the line passing through (1,2,3) and whose D.R.'s are 2,3,6,
x12=y+23=z36=r, say.
Any point on it is (2r+1,3r2,6r+3).
If it lies on the plane xy+z=5, then
(2r+1)(3r2)+(6r+3)=5.
r=17
The point is (97,117,157) and its distance from (1,2,3)

is 17(4+9+36)=77=1.

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