# JEE-Mains 2015 (23/30)

Geometry Level 3

The distance of the point $$(1,0,2)$$ from the point of intersection of the line $$\dfrac{x-2}{3}=\dfrac{y+1}{4}=\dfrac{z-2}{12}$$ and the plane $$x-y+z=16$$ is :

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