Let the line x/1=(6-y)/2=(z+8)/5 intersect the lines (x-5)/4=(y-7)/3=(z+2)/1 and (x+3)/6=(3-y)/3=(z-6)/1 at the points A and B respectively. Then the…

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