3d Geometry

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 \(\mathrm{AB}\) from the plane \(2 x-2 y+z=14\) is :