The distance, of the point $(7,-2,11)$ from the line $\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is:

$B=(2\lambda +7,-3\lambda -2,6\lambda +11)$

Point B lies on  
$$\begin{gathered} \frac{2\lambda +7-6}{1}=\frac{-3\lambda -2-4}{0}=\frac{6\lambda +11-8}{3} \\ -3\lambda -6=0 \lambda =-2 \\ B\Rightarrow (3,4,-1) \\ AB=\sqrt{{(7-3)}^2+{(4+2)}^2+{(11+1)}^2}=\sqrt{196}=14 \end{gathered}$$