Find the distance of the point $( - 1, - 5, - 10)$ from the point of intersection of the line $\vec r = 2\hat i - \hat j + 2\hat k + \lambda (3\hat i + 4\hat j + 2\hat k)$and the plane $\vec r \cdot (\hat i - \hat j + \hat k) = 5.$

.: The given line is $\vec r = 2\hat i - \hat j + 2\hat k + \lambda (3\hat i + 4\hat j + 2\hat k)$ ...(1)

and the given plane is$\vec r \cdot (\hat i - \hat j + \hat k) = 5$ ...(2)

Solving (1) and (2),

we get
$[2\hat i - \hat j + 2\hat k + \lambda (3\hat i + 4\hat j + 2\hat k)] \cdot (\hat i - \hat j + \hat k) = 5$

$\Rightarrow$ $\left( {(2 + 3\lambda )\hat i + (4\lambda - 1)\hat j + (2\lambda + 2)\hat k} \right) \cdot (\hat i - \hat j + \hat k) = 5$

$\Rightarrow$ $(2 + 3\lambda )(1) + (4\lambda - 1)( - 1) + (2\lambda + 2)(1) = 5$

$\Rightarrow$ $2 + 3\lambda - 4\lambda + 1 + 2\lambda + 2 = 5 \Rightarrow \lambda = 0$

$\therefore$ The point of intersection of (1) and (2) is

$(2\hat i - \hat j + 2\hat k)$ i.e, $(2, - 1,2)$

The other point is $( - 1, - 5, - 10)$

$\therefore$ Required distance $= \sqrt {{{( - 1 - 2)}^2} + {{( - 5 + 1)}^2} + {{( - 10 - 2)}^2}}$

$= \sqrt {9 + 16 + 144} = \sqrt {169} = 13$ units.