Answer:
Option C
Explanation:
Direction ratio of line perpendicular to the lines $\overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$
and $\overrightarrow{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$ is $\alpha(\overrightarrow{b_{1}}\times\overrightarrow{b_{2}})$
$\therefore$ Direction ratio of line perpendicular to the lines
$\overrightarrow{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2\hat{i}-2\hat{j}+2\hat{k})$
and
$\overrightarrow{r}=(2\hat{i}+\hat{j}-3\hat{k})+\mu(\hat{i}-2\hat{j}+2\hat{k})$ is
$\alpha\begin{bmatrix}\hat{i} & \hat{j}&\hat{k} \\2 & -2&1 \\1 &-2 &2 \end{bmatrix}$
= $\alpha [{(-4+2)\hat{i}-(4-1)\hat{j}+(-4+2)\hat{k}]}$
= $\alpha [{-2\hat{i}-3\hat{j}-2\hat{k}]}$
Now, equation of line passing through (3,-1,2) and parallel to ${-2\hat{i}-3\hat{j}-2\hat{k}}$
$\overrightarrow{r}=3\hat{i}-\hat{j}-2\hat{k}+\beta(2\hat{i}+3\hat{j}+2\hat{k})$
Hence, cartesian form of the above equation is
$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$
