Let the distance between plane passing through the lines 
$\frac{\mathrm{x}+1}{2}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}+1}{8}\text{ and }\frac{\mathrm{x}+3}{2}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\mathrm{\lambda }}$ and the plane $23\mathrm{x}-10\mathrm{y}-2\mathrm{z}+48=0$ be $\frac{\mathrm{k}}{\sqrt{633}}\text{. Then }2\mathrm{k}+3\mathrm{\lambda }=$

The distance between the plane containing the given lines and the plane 23x -10y - 2z + 48 = 0 is equal to the length of the perpendicular from any point on one of the given lines to the plane 23x -10y - 2z + 48 = 0. The coordinates of a point on first line are (-1, 3, -1). Length of the perpendicular form (-1, 3, -1) to the plane 23x -10y - 2z + 48 = 0 is
$\begin{array}{l}\frac{|-23-30+2+48|}{\sqrt{529+100+4}}=\frac{3}{\sqrt{633}}\\ \therefore \frac{\mathrm{k}}{\sqrt{633}}=\frac{3}{\sqrt{633}}\Rightarrow \mathrm{k}=3\end{array}$
Given lines are coplanar.
$\begin{array}{l}\therefore \left|\begin{array}{ccc}-3+1& -2-3& 1+1\\ 2& 2& 8\\ 2& 1& \mathrm{\lambda }\end{array}\right|=0\\ \Rightarrow -2(2\mathrm{\lambda }-8)+5(2\mathrm{\lambda }-16)+2(2-4)=0\\ \Rightarrow -4\mathrm{\lambda }+16+10\mathrm{\lambda }-80-4=0\Rightarrow 6\mathrm{\lambda }-68=0\Rightarrow \mathrm{\lambda }=\frac{34}{3}\end{array}$