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Q.
The distance of the point of intersection of the line $ \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12} $ and me plane $ x-y+z=5 $ from the point $ (-1,-5,-10) $ is
Given line is $ \frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=k(say) $ Any point on the line is $ (3k+2,4k-1,12k+2) $ This point lies on the plane $ x-y+z=5 $
$ \therefore $ $ 3k+2-(4k-1)+12k+2=5 $
$ \Rightarrow $ $ 11k=0 $
$ \Rightarrow $ $ k=0 $
$ \therefore $ Intersection point is $ (2,-1,2) $ .
$ \therefore $ Distance, between points $ (2,-1,2) $ and $ (-1,-5,-10) $
$=\sqrt{{{(-1-2)}^{2}}+{{(-5+1)}^{2}}+{{(-10-2)}^{2}}} $
$=\sqrt{9+16+144}=13 $