Given lines are $\frac{1-\mathit{x}}{3}$ = $\frac{\mathit{y}-2}{2\mathit{\lambda }}$ = $\frac{\mathit{z}-3}{2}$ and $\frac{\mathit{x}-1}{3\mathit{\lambda }}$ = $\frac{\mathit{y}-1}{1}$ = $\frac{6-\mathit{z}}{7}$
Let us rewrite the equations of the given lines as follows:
$\frac{-\left(\mathit{x}-1\right)}{3}$ = $\frac{\mathit{y}-2}{2\mathit{\lambda }}$ = $\frac{\mathit{z}-3}{2}$ and $\frac{\mathit{x}-1}{3\mathit{\lambda }}$ = $\frac{\mathit{y}-1}{1}$ = $\frac{-\left(\mathit{z}-6\right)}{7}$
That is we have,
$\frac{\mathit{x}-1}{-3}$ = $\frac{\mathit{y}-2}{2\mathit{\lambda }}$ = $\frac{\mathit{z}-3}{2}$ and $\frac{\mathit{x}-1}{3\mathit{\lambda }}$ = $\frac{\mathit{y}-1}{1}$ = $\frac{\mathit{z}-6}{-7}$
The lines are perpendicular so angle between them is 90
So, cos θ = 0
Here $\left({\mathit{a}}_1,{\mathit{b}}_1,{\mathit{c}}_1\right)$ = (-3,2λ,2) and $\left({\mathit{a}}_2,{\mathit{b}}_2,{\mathit{c}}_2\right)$ = (3λ,1,-7)
For perpendicular lines
${\mathit{a}}_1{\mathit{a}}_2+{\mathit{b}}_1{\mathit{b}}_2+{\mathit{c}}_1{\mathit{c}}_2$ = 0
⇒ - 9λ + 2λ - 14 = 0
⇒ - 7λ = 14
⇒ λ = $\frac{14}{-7}$
⇒ λ = - 2