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