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