Question:
If the distance between the plane, $23 x-10 y-2 z+48=0$ and the plane
containing the lines $\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}$ and $\frac{\mathrm{x}+3}{2}=\frac{\mathrm{y}+2}{6}=\frac{\mathrm{z}-1}{\lambda}(\lambda \in \mathrm{R})$
is equal to $\frac{\mathrm{k}}{\sqrt{633}}$, then $\mathrm{k}$ is equal to__________.
Solution:
