Let d be the distance of the point of intersection of the lines...

$$\begin{aligned} & \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda....(1)\\ & x=3 \lambda-6, y=2 \lambda, z=\lambda-1 \\ & \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu....(2)\end{aligned}$$\mathrm{x}=4 \mu+7, \mathrm{y}=3 \mu+9, \mathrm{z}=2 \mu+4 3 \lambda-6=4 \mu+7 \Rightarrow 3 \lambda-4 \mu=13 \ldots(3) \times 2 2 \lambda=3 \mu+9 \Rightarrow 2 \lambda-3 \mu=9 \ldots(4) \times 3 $$\begin{array}{c} 6 \lambda-8 \mu=26 \\ 6 \lambda-9 \mu=27 \\ -\quad+\quad- \\ \hline\mu=-1 \end{array}$$ \begin{gathered} \Rightarrow 3 \lambda-4(-1)=13 \\ 3 \lambda=9 \\ \lambda=3 \end{gathered}$int. point$(3,6,2) ;(7,8,9) \mathrm{d}^2=$16+4+49=69 Ans.$\mathrm{d}^2+6=69+6=75$