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JEEMAIN-2015
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The distance of the point $(1,0,2)$ from the point of intersection of the line $\large\frac{x-2}{3} =\frac{y+1}{4} =\frac{z-2}{12}$ and the plane $x-y+z=16,$ is :
$\begin{array}{1 1} 13 \\ 2 \sqrt {14} \\ 8 \\ 3 \sqrt{21} \end{array} $
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