Solution
line $\mathrm{L}_{1} \frac{\mathrm{x}-2}{-3}=\frac{\mathrm{y}-6}{2}=\frac{\mathrm{z}-7}{4}=\lambda_{1}$
line $\mathrm{L}_{2} \frac{\mathrm{x}-4}{2}=\frac{\mathrm{y}-3}{1}=\frac{\mathrm{z}-5}{3}=\lambda_{2}$
Point $\mathrm{C}\left(-3 \lambda_{1}+2,2 \lambda_{1}+6,4 \lambda_{1}+7\right)$ on line $\mathrm{L}_{1}$
Point $\mathrm{D}\left(2 \lambda_{2}+4, \lambda_{2}+3,3 \lambda_{2}+5\right)$ on line $\mathrm{L}_{2}$
\[
\begin{aligned}
& \mathrm{CD} \xrightarrow{\mathrm{drs}} 2 \lambda_{2}+3 \lambda_{1}+2, \lambda_{2}-2 \lambda_{1}-3,3 \lambda_{2}-4 \lambda_{1}-2 \\
& \text { or }-3,5,16 \\
& \frac{2 \lambda_{2}+3 \lambda_{1}+2}{-3}=\frac{\lambda_{2}-2 \lambda_{1}-3}{5}=\frac{3 \lambda_{2}-4 \lambda_{1}-2}{16}
\end{aligned}
\]
From first \& middle
\[
\begin{aligned}
& 10 \lambda_{2}+15 \lambda_{1}+10=-3 \lambda_{2}+6 \lambda_{1}+9 \\
& \Rightarrow 13 \lambda_{2}+9 \lambda_{1}+1=0
\end{aligned}
\]
From middle \& third
\[
\begin{aligned}
& 16 \lambda_{2}-32 \lambda_{1}-48=15 \lambda_{2}-20 \lambda_{1}-10 \\
& \Rightarrow \lambda_{2}=12 \lambda_{1}+38 \\
& 13\left(12 \lambda_{1}+38\right)+9 \lambda_{1}+1=0 \\
& \Rightarrow 165 \lambda_{1}+495=0 \Rightarrow \lambda_{1}=-3 \\
& \Rightarrow \lambda_{2}=-36+38 \Rightarrow \lambda_{2}=2 \text { Ans. } \\
& \lambda_{1}=-3, \lambda_{2}=2
\end{aligned}
\]
we get $\mathrm{C}(11,0,-5) \mathrm{D}(8,5,11)$
\[
\mathrm{CD}^{2}=9+25+256=290 \text { Answer }(2)
\]