If the distance between the plane

Since, the line $\frac{x+1}{2}=\frac{y-3}{4}=\frac{z+1}{3}$ contains the point

$(-1,3,-1)$ and line $\frac{x+3}{2}=\frac{y+2}{6}=\frac{z-1}{\lambda}$ contains the

point $(-3,-2,1)$

Then, the distance between the plane

$23 x-10 y-2 z+48=0$ and the plane containing the

lines $=$ perpendicular distance of plane

$23 x-10 y-2 z+48=0$ either from $(-1,3,-1)$ or $(-3,-2,1)$

$=\left|\frac{23(-1)-10(3)-2(-1)}{\sqrt{(23)^{2}+(10)^{2}+(-2)^{2}}}\right|=\frac{3}{\sqrt{633}}$

It is given that distance between the planes

$=\frac{k}{\sqrt{633}} \Rightarrow \frac{k}{\sqrt{633}}=\frac{3}{\sqrt{633}} \Rightarrow k=3$