Question

Find the equation of a line passing through the point $(-1,3,-2)$ and perpendicular to the lines: $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ and $\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$.

Answer 1

Any line through the point $(-1,3,-2)$ is
$\frac{x-(-1)}{a}=\frac{y-3}{b}=\frac{z-(-2)}{c}$ .......$\left(i\right)$

It is perpendicular to the lines
$\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ ......$\left(ii\right)$
and

$\frac{x+2}{-3}=\frac{y-1}{2}=\frac{z+1}{5}$ ....$\left(iii\right)$
then $1.a+2.b+3.c=0$
$\Rightarrow a+2b+3c=0$ ....... $\left(iv\right)$
and $-3.a+2.b+5.c=0$
$\Rightarrow -3a+2b+5c=0$ ...... $\left(v\right)$

Substracting $\left(v\right)$ from $\left(iv\right)$, we get,
$4a-2c=0$
$\Rightarrow c=2a$
and hence from (iv)
$a+2b+3.2a=0$
$\Rightarrow 2b=-7a$
$\Rightarrow b=\frac{-7a}{2}$ ........ $\left(vii\right)$

From (vi) and (vii)
$a:b:c=a:-\frac{7a}{2}:2a$
$\Rightarrow a:b:c=2:-7:4$

Putting these values in (i), the equation of the required lines are
$\frac{x+1}{2}=\frac{y-3}{-7}=\frac{z+2}{4}$