The distance of the point $(5,3,-1)$ from the plane passing through points $(2,1,0),(3,-2,4)$ and $(1,-3,3)$ is

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

A line L is passing through points $\mathrm{A}(1,3,2)$ and $\mathrm{B}(2,2,1)$. If mirror image of point $\mathrm{P}(1,1,-1)$ in the line L is $(x, y, z)$ then $x+y+\mathrm{z}=$

The lines $\frac{6 x-6}{18}=\frac{y+1}{3}=\frac{z-1}{5} \quad$ and $\frac{3 x+6}{12}=\frac{y-1}{3}=\frac{z+1}{2}$ are $\ldots$