$$\begin{aligned} & L_1: \frac{x-2}{1}=\frac{y}{-1}=\frac{z-1}{1}=\lambda \\ & L_2: \frac{x+1}{(1 / 2)}=\frac{y-1}{(1 / 2)}=\frac{z+1}{1}=\mu \end{aligned}$$

Any point of $$L_1$$ and $$L_2$$ will be $$(\lambda+2,-\lambda, \lambda+1)$$ and $$\left(\frac{\mu}{2}-1, \frac{\mu}{2}+1, \mu-1\right)$$

Now Dr of line $$<\lambda-\frac{\mu}{2}+3,-\lambda-\frac{\mu}{2}-1, \lambda-\mu+2>$$

Now $$\frac{\lambda-\frac{\mu}{3}+3}{3}=\frac{-\lambda-\frac{\mu}{2}-1}{1}=\frac{\lambda-\mu+2}{2}$$

$$\left. {\matrix{ {\lambda - {\mu \over 3} + 3 = - 3\lambda - {{3\mu } \over 2} - 3\,\,\,...(1)} \cr {2\left( {\lambda - {\mu \over 3} + 3} \right) = 3(\lambda - \mu + 2)\,\,...(2)} \cr } } \right\}\lambda = {{ - 4} \over 3},\mu = {{ - 2} \over 3}$$

$$\therefore \quad$$ Points will be $$\left(\frac{2}{3}, \frac{4}{3}, \frac{-1}{3}\right)$$ and $$\left(\frac{-4}{3}, \frac{2}{3}, \frac{-5}{3}\right)$$

$$\therefore \quad L$$ will be $$\frac{x-\frac{2}{3}}{3}=\frac{y-\frac{4}{3}}{1}=\frac{z+\frac{1}{3}}{2}$$

$$\therefore \quad\left(\frac{-1}{3}, 1,-1\right)$$ will satisfy $$L$$