If the line drawn from the point $\left( { - 2, - 1, - 3} \right)$ meets a plane at right angle at the point $(1, - 3,3)$, then find the equation of the plane.

Since, the line drawn from the point (-2,-1,-3) meets a plane at right angle at the point $(1, - 3,3)$, So,

the plane passes through the point $(1, - 3,3)$ and normal to plane is $( - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} - 6\widehat {\rm{k}})$.

$\Rightarrow$ $\overrightarrow {\rm{a}} = \widehat {\rm{i}} - 3\widehat {\rm{j}} + 3\widehat {\rm{k}}$

and $\overrightarrow {\rm{N}} = - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} - 6\widehat {\rm{k}}$

So, the equation of required plane is $(\overrightarrow {\rm{r}} - \overrightarrow {\rm{a}} ) \cdot \overrightarrow {\rm{N}} = 0$

$\Rightarrow$ $[(x\widehat {\rm{i}} + y\widehat {\rm{j}} + z\widehat {\rm{k}}) - (\widehat {\rm{i}} - 3\widehat {\rm{j}} + 3\widehat {\rm{k}})] \cdot ( - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} - 6\widehat {\rm{k}}) = 0$

$\Rightarrow$ $[(x - 1)\widehat {\rm{i}} + (y + 3)\widehat {\rm{j}} + (z - 3)\widehat {\rm{k}}] \cdot ( - 3\widehat {\rm{i}} + 2\widehat {\rm{j}} - 6\widehat {\rm{k}}) = 0$

$\Rightarrow$ $- 3x + 3 + 2y + 6 - 6z + 18 = 0$

$\Rightarrow$ $- 3x + 2y - 6z = - 27$
$\therefore$ $3x - 2y + 6z - 27 = 0$