The equations of a line passing through $(3,-1,2)$ and perpendicular to the lines…
The equations of a line passing through $(3,-1,2)$ and perpendicular to the lines $\overline{\mathrm{r}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})+\lambda(2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})$ and $\bar{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\mu(\hat{i}-2 \hat{j}+2 \hat{k})$ is
$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$
$\frac{x-3}{3}=\frac{y+1}{2}=\frac{z-2}{2}$
$\frac{x+3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$
$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{3}$
Solution
Let $a, b, c$ be the direction ratios of the required line.
$\therefore 2 \mathrm{a}-2 \mathrm{~b}+\mathrm{c}=0$
... (1) and $a-2 b+2 c=0$
From (1) and (2), we write
$\begin{aligned}
& \frac{\mathrm{a}}{\left|\begin{array}{ll}
-2 & 1 \\
-2 & 1
\end{array}\right|}=\frac{\mathrm{b}}{\left|\begin{array}{ll}
2 & 1 \\
1 & 2
\end{array}\right|}=\frac{\mathrm{c}}{\left|\begin{array}{ll}
2 & -2 \\
1 & -2
\end{array}\right|} \quad \Rightarrow \frac{\mathrm{a}}{-2}=\frac{\mathrm{b}}{-3}=\frac{\mathrm{c}}{-2} \\
& \therefore(\mathrm{a}, \mathrm{b}, \mathrm{c})=(2,3,2)
\end{aligned}$
So equation of required line is
$\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-2}{2}$