The foot of a perpendicular drawn from the point (-2, - 1, -3) on a plane is (1, -3, 3). Find the equation of the plane.

From the given data, the line drawn from the point (-2, -1, -3) meets a plane at right angle at the point (1, -3, 3). hence, the plane passes through the point (1, -3,3) and normal to plane is$(-3\hat{\mathrm{i}}+2\hat{\mathrm{j}}-6\hat{\mathrm{k}}).$
$\begin{array}{l}\Rightarrow \overrightarrow{\mathrm{a}}=\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\\ \text{ and }\overrightarrow{\mathrm{N}}=-3\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}}\end{array}$
Therefore, the equation of asked plane is
$\begin{array}{l}(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}})\cdot \overrightarrow{\mathrm{N}}=0\\ \Rightarrow \left[\right(\mathrm{x}\widehat{\mathrm{i}}+\mathrm{y}\widehat{\mathrm{j}}+\mathrm{z}\widehat{\mathrm{k}})-(\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+3\widehat{\mathrm{k}}\left)\right]\cdot (-3\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}})=0\\ \Rightarrow \left[\right(\mathrm{x}-1)\widehat{\mathrm{i}}+(\mathrm{y}+3)\widehat{\mathrm{j}}+(\mathrm{z}-3\left)\widehat{\mathrm{k}}\right)]\cdot (-3\widehat{\mathrm{i}}+2\widehat{\mathrm{j}}-6\widehat{\mathrm{k}})=0\\ \Rightarrow -3\mathrm{x}+3+2\mathrm{y}+6-6\mathrm{z}+18=0\\ \Rightarrow -3\mathrm{x}+2\mathrm{y}-6\mathrm{z}=-27\\ \therefore 3\mathrm{x}-2\mathrm{y}+6\mathrm{z}-27=0\end{array}$
Therefore the equation of the plane is 3x – 2y + 6z + 27 = 0.