The equation of the line passing through the point $(-1,3,-2)$ and perpendicular to each of the lines…
The equation of the line passing through the point $(-1,3,-2)$ and perpendicular to each of the lines $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$ is
$\frac{x+1}{2}=\frac{y-3}{7}=\frac{z+2}{4}$
$\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$
$\frac{x-1}{2}=\frac{y+3}{7}=\frac{z-2}{4}$
$\frac{x-1}{2}=\frac{y+3}{-7}=\frac{z-2}{4}$
Solution
The direction ratios of the required line can be obtained by
$\begin{aligned} & \frac{\mathrm{a}}{2 \times 5-2 \times 3}=\frac{\mathrm{b}}{-3 \times 3-1 \times 5}=\frac{\mathrm{c}}{1 \times 2-(-3) \times 2} \\ & \Rightarrow\langle\mathrm{a}, \mathrm{b}, \mathrm{c}\rangle \equiv\langle 4,-14,8\rangle \equiv\langle 2,-7,4\rangle\end{aligned}$
Hence, the equation of required line
$\begin{aligned} & \frac{x-(-1)}{2}=\frac{y-3}{-7}=\frac{z-(-2)}{4} \\ & \Rightarrow \frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}\end{aligned}$