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

Given,

The line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$  and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $\mathrm{A}$ and $\mathrm{B}$ respectively,

Now plotting the diagram, we get,

Now solving, $\frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\lambda$ and $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}=t$ we get,

$A\equiv (\lambda ,-2\lambda +6,5\lambda -8)\equiv (4t+5,3t+7,t-2)$

Now comparing both side and solving we get, $t=-1,\lambda =1$

Hence, point$A(1, 4, –3)$

Now solving, $\frac{x}{1}=\frac{y-6}{-2}=\frac{z+8}{5}=\mu$ and $\frac{x+3}{6}=\frac{y-3}{-3}=\frac{z-6}{1}=k$ we get,

For point $B\equiv (\mu ,-2\mu +6,5\mu -8)$ $\equiv (6k-3,-3k+3,k+6)$

Now comparing both side and solving we get, $k=1,\mu =3$

Hence, point $B(3, 0, 7)$

So, mid-point of $AB=M(2, 2, 2)$

Now finding, distance of $M$ from the plane $2x – 2y + z – 14 = 0$ we get,

$\left|\frac{4-4+2-14}{\sqrt{2^2+2^2+1^2}}\right|=\left|\frac{{\displaystyle -12}}{{\displaystyle 3}}\right|=4$