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}…

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 \(A\) and \(B\) respectively. Then the distance of the midpoint of the line segment \(A B\) from the plane \(2 x-2 y+z\) \(=14\) is

[JEE Main 2023, 10 Apr (Shift 2)]