The equation of the line through the point $(0,1,2)$ and perpendicular to the line $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{-2}$ is

If lines $\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2}$ and $\frac{x-1}{3 k}=\frac{y-5}{1}=\frac{z-6}{-5}$ are mutually perpendicular, then $k$ is equal to

The distance between the two planes $2 x+3 y+4 z=4$ and $4 x+6 y+8 z=12$ is