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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 :

3\sqrt{21}

13

2\sqrt{14}

8

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by Sandeep Bhardwaj

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