Equation of line is $\overrightarrow{\mathit{r}}$ = $2\hat{\mathit{i}}-\hat{\mathit{j}}+2\hat{\mathit{k}}$ + λ $\left(3\hat{\mathit{i}}+4\hat{\mathit{j}}+2\hat{\mathit{k}}\right)$
$\overrightarrow{\mathit{r}}$ = (2 + 3λ) $\hat{\mathit{i}}$ + (- 1 + 4λ) $\hat{\mathit{j}}$ + (2 + 2λ) $\hat{\mathit{k}}$
This lies on the plane $\overrightarrow{\mathit{r}}$ = $\left(\hat{\mathit{i}}-\hat{\mathit{j}}+\hat{\mathit{k}}\right)$ = 5
⇒ [(2 + 3λ) $\hat{\mathit{i}}$ + (- 1 + 4λ) $\hat{\mathit{j}}$ + (2 + 2λ) $\hat{\mathit{k}}$] . $\left(\hat{\mathit{i}}-\hat{\mathit{j}}+\hat{\mathit{k}}\right)$ = 5
⇒ (2 + 3λ) + (- 1 + 4λ) (- 1) + (2 + 2λ) = 5
⇒ λ = 0
Coordinates are
(2 + 3λ , - 1 + 4λ , 2 + 2λ) = (2 , - 12)
Distance between (2 , - 1 , 2) and (- 1 , - 5 , - 10)
=
= 13 units