The distance of the point $(1,3,-7)$ from the plane passing through the point $(1,-1,-1)$ having normal perpendicular to both the lines $\frac{x-1}{1}=\frac{y+2}{-2}=\frac{z-4}{3}$ and $\frac{x-2}{2}=\frac{y+1}{-1}=\frac{z+7}{-1}$ is

The value of m , such that $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-m}{2}$ lies in the plane $2 x-4 y+z=7$, is

The length of the perpendicular from the point $\mathrm{A}(1,-2,-3)$ on the line $\frac{x-1}{2}=\frac{y+3}{-1}=\frac{z+1}{-2}$ is

If the points $(1,-1, \lambda)$ and $(-3,0,1)$ are equidistant from the plane $3 x-4 y-12 z+13=0$, then the sum of all possible values of $\lambda$ is