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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} 3 \sqrt {21} \\ 13 \\ 2 \sqrt {14} \\ 8 \end{array} $

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