Let $\overrightarrow{{\mathit{b}}_1}$ and $\overrightarrow{{\mathit{b}}_2}$ be the vector parallel to the pair to lines,
$\frac{-\mathit{x}+2}{-2}$ = $\frac{\mathit{y}-1}{7}$ = $\frac{\mathit{z}+3}{-3}$ and $\frac{\mathit{x}+2}{-1}$ = $\frac{2\mathit{y}-8}{4}$ = $\frac{\mathit{z}-5}{4}$ , respectively.
Now, $\frac{-\mathit{x}+2}{-2}$ = $\frac{\mathit{y}-1}{7}$ = $\frac{\mathit{z}+3}{-3}$ ⇒ $\frac{\mathit{x}-2}{2}$ = $\frac{\mathit{y}-1}{7}$ = $\frac{\mathit{z}+3}{-3}$
$\frac{\mathit{x}+2}{-1}$ = $\frac{2\mathit{y}-8}{4}$ = $\frac{\mathit{z}-5}{4}$
⇒ $\frac{\mathit{x}+2}{-1}$ = $\frac{\mathit{y}-4}{2}$ = $\frac{\mathit{z}-5}{4}$
∴ $\overrightarrow{{\mathit{b}}_1}$ = $2\hat{\mathit{i}}+7\hat{\mathit{j}}-3\hat{\mathit{k}}$ and $\overrightarrow{{\mathit{b}}_2}$ = $-\hat{\mathit{i}}+2\hat{\mathit{j}}+4\hat{\mathit{k}}$
$\left|\overrightarrow{{\mathit{b}}_1}\right|$ = $\sqrt{{\left(2\right)}^2+{\left(7\right)}^2+{\left(-3\right)}^2}$ = $\sqrt{62}$
$\left|\overrightarrow{{\mathit{b}}_2}\right|$ = $\sqrt{{\left(-1\right)}^2+{\left(2\right)}^2+{\left(4\right)}^2}$ = $\sqrt{21}$
$\overrightarrow{{\mathit{b}}_1}.\overrightarrow{{\mathit{b}}_2}$ =
= 2 (- 1) + 7 × 2 + (- 3) . 4
= - 2 + 14 - 12
= 0
The angle θ between the given pair of lines is given by the relation,
cos θ = $\left|\frac{\overrightarrow{{\mathit{b}}_1}.\overrightarrow{{\mathit{b}}_2}}{\left|\overrightarrow{{\mathit{b}}_1}\right||\overrightarrow{{\mathit{b}}_2}|}\right|$
⇒ cos θ = $\frac{0}{\sqrt{62}\times \sqrt{21}}$ = 0
⇒ θ = $\mathit{c}\mathit{o}{\mathit{s}}^{-1}$ (0) = $\frac{\mathit{\pi }}{2}$
Thus, the given lines are perpendicular to each other and the angle between them is 90°.