We know that, equation of a line passing through ${\mathit{x}}_1,{\mathit{y}}_1,{\mathit{z}}_1$ with direction ratios a, b, c
$\frac{\mathit{x}-{\mathit{x}}_1}{\mathit{a}}$ = $\frac{\mathit{y}-{\mathit{y}}_1}{\mathit{b}}$ = $\frac{\mathit{z}-{\mathit{z}}_1}{\mathit{c}}$
So, the required equation of a line passing through ( 1,3, 2) is:
$\frac{\mathit{x}+1}{\mathit{a}}$ = $\frac{\mathit{y}-3}{\mathit{b}}$ = $\frac{\mathit{z}+2}{\mathit{c}}$ ... (1)
Given that line $\frac{\mathit{x}}{1}$ = $\frac{\mathit{y}}{2}$ = $\frac{\mathit{z}}{3}$ is perpendicular to line (1),so
${\mathit{a}}_1{\mathit{a}}_2+{\mathit{b}}_1{\mathit{b}}_2+{\mathit{c}}_1{\mathit{c}}_2$ = 0
a + 2b + 3c = 0 ... (2)
And line $\frac{\mathit{x}+2}{-3}$ = $\frac{\mathit{y}-1}{2}$ = $\frac{\mathit{z}+1}{5}$ is perpendicular to line 1 , so
${\mathit{a}}_1{\mathit{a}}_2+{\mathit{b}}_1{\mathit{b}}_2+{\mathit{c}}_1{\mathit{c}}_2$ = 0
- 3a + 2b + 5c = 0 ... (3)
Solving equation 2 and 3 by cross multiplication
$\frac{\mathit{a}}{\left(2\right)\left(5\right)-\left(2\right)\left(3\right)}$ = $\frac{\mathit{b}}{\left(-3\right)\left(3\right)-\left(1\right)\left(5\right)}$ = $\frac{\mathit{c}}{\left(1\right)\left(2\right)-\left(-3\right)\left(2\right)}$
⇒ $\frac{\mathit{a}}{10-6}$ = $\frac{\mathit{b}}{-9-5}$ = $\frac{\mathit{c}}{2+6}$
⇒ $\frac{\mathit{a}}{4}$ = $\frac{\mathit{b}}{-14}$ = $\frac{\mathit{c}}{8}$
⇒ $\frac{\mathit{a}}{2}$ = $\frac{\mathit{b}}{-7}$ = $\frac{\mathit{c}}{4}$ = λ (say)
⇒ a = 2λ , b = - 7 λ , c = 4λ
Putting the value of a,b, and c in (1) gives
$\frac{\mathit{x}+1}{2\mathit{\lambda }}$ = $\frac{\mathit{y}-3}{-7\mathit{\lambda }}$ = $\frac{\mathit{z}+2}{4\mathit{\lambda }}$
⇒ $\frac{\mathit{x}+1}{2}$ = $\frac{\mathit{y}-3}{-7}$ = $\frac{\mathit{z}+2}{4}$