The distance of the point $$(1,6,2)$$ 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

A line drawn from the point $$\mathrm{A}(1,3,2)$$ parallel to the line $$\frac{x}{2}=\frac{y}{4}=\frac{z}{1}$$, intersects the plane $$3 x+y+2 z=5$$ in point $$\mathrm{B}$$, then co-ordinates of point $$\mathrm{B}$$ are

A line $$\mathrm{L}_1$$ passes through the point, whose p. v. (position vector) $$3 \hat{i}$$, is parallel to the vector $$-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$$. Another line $$\mathrm{L}_2$$ passes through the point having p.v. $$\hat{i}+\hat{j}$$ is parallel to vector $$\hat{i}+\hat{k}$$, then the point of intersection of lines $$L_1$$ and $$L_2$$ has p.v.

The equation of the line passing through the point $$(-1,3,-2)$$ and perpendicular to each of the lines $$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$$ and $$\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$$ is