The distance of a point (2, 5, –3) from the plane  $r\cdot (6\widehat{i}-3\widehat{j}+2\widehat{k})=4$ is

Here $a=2\widehat{i}+5\widehat{j}-3\widehat{k},N=6\widehat{i}-3\widehat{j}+2\widehat{k}\text{ and }d=4$.

Therefore, the distance of the point (2, 5, –3) from the

given plane is $\frac{\left|\right(2\widehat{\mathrm{i}}+5\widehat{\mathrm{j}}-3\widehat{\mathrm{k}})\cdot (6\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+2\widehat{\mathrm{k}})-4|}{|6\widehat{\mathrm{i}}-3\widehat{\mathrm{j}}+2\widehat{\mathrm{k}}|}$

$=\frac{|12-15-6-4|}{\sqrt{36+9+4}}=\frac{13}{7}\left(\because \text{ distance }=\mid \frac{a\cdot N-d}{\overline{N}}\right)$