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
Let $\mathrm{a}, \mathrm{b}, \mathrm{c}$ be the direction ratios of the required line.
Since the line is perpendicular to the lines with d.r.s. 1, 2, 3 and $-3,2,5$.
$\begin{array}{ll}
\therefore \quad & a+2 b+3 c=0 \\
& \text { and }-3 a+2 b+5 c=0 \\
& \Rightarrow \frac{a}{2}=\frac{b}{-7}=\frac{c}{4}
\end{array}$
...[From (i) and (ii)]
$\therefore \quad$ Equation of the required line is
$\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$