Let the line $\frac{\mathrm{x}}{1}=\frac{6-\mathrm{y}}{2}=\frac{\mathrm{z}+8}{5}$ intersects the line $\frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-7}{3}=\frac{\mathrm{z}+2}{1}$ and $\frac{\mathrm{x}+3}{6}=\frac{3-\mathrm{y}}{3}=\frac{\mathrm{z}-6}{1}$ at the points A and B respectively. Then the distance of the mid point of the line segment AB from the plane 2x-2y+z=14 is

$$\begin{gathered} \text{ Point of intersection of lines }\frac{\mathrm{x}}{1}=\frac{6-\mathrm{y}}{2}=\frac{\mathrm{z}+8}{5}\text{ and }\frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-7}{3}=\frac{\mathrm{z}+2}{1} \\ \begin{array}{l}\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-6}{-2}=\frac{\mathrm{z}+8}{5}=\mathrm{\lambda }\text{ and }\frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-7}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k}\\ \mathrm{x}=\mathrm{\lambda }=4\mathrm{k}+5|\mathrm{y}=-2\mathrm{\lambda }+6=3\mathrm{k}+7|\mathrm{z}=5\mathrm{\lambda }-8=\mathrm{k}-2\\ \Rightarrow \mathrm{k}=-1,\mathrm{\lambda }=1\text{ and }\mathrm{A}=(1,4,-3)\end{array} \\ \text{ Point of intersection of lines }\frac{\mathrm{x}}{1}=\frac{6-\mathrm{y}}{2}=\frac{\mathrm{z}+8}{5}\text{ and }\frac{\mathrm{x}+3}{6}=\frac{3-\mathrm{y}}{3}=\frac{\mathrm{z}-6}{1} \\ \begin{array}{l}\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-6}{-2}=\frac{\mathrm{z}+8}{5}=\mathrm{\lambda }\text{ and }\frac{\mathrm{x}+3}{6}=\frac{\mathrm{y}-3}{-3}=\frac{\mathrm{z}-6}{1}=\mathrm{k}\\ \mathrm{x}=\mathrm{\lambda }=6\mathrm{k}-3|\mathrm{y}=-2\mathrm{\lambda }+6=-3\mathrm{k}+3|\mathrm{z}=5\mathrm{\lambda }-8=\mathrm{k}+6\\ \Rightarrow \mathrm{k}=1,\mathrm{\lambda }=3\text{ and }\mathrm{B}=(3,0,7)\\ \text{ Midpoint }\mathrm{C}=(2,2,2)\\ \text{ Distance of C from the plane }2\mathrm{x}-2\mathrm{y}+\mathrm{z}=14\text{ is }\left|\frac{4-4+2-14}{\sqrt{4+4+1}}\right|=4\end{array} \end{gathered}$$