Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

(B) Let $P(3,2,1)$ be the foot of the perpendicular from the origin to the plane. $\begin{array}{rl}\overline{\mathrm{OP}}& =\overline{\mathrm{n}}=3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}\text{ and }\\ \mathrm{P}& =|3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}|=\sqrt{3^2+2^2+1^2}=\sqrt{14}\\ \hat{\mathrm{n}}& =\frac{3\hat{\mathrm{i}}+2\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{14}}\end{array}$ The vector equation of the plane is $\begin{array}{rl}& (x\hat{i}+y\hat{j}+z\hat{k})\cdot \frac{(3\hat{i}+2\hat{j}+\hat{k})}{\sqrt{14}}=\sqrt{14}\\ \therefore & 3x+2y+z=14\Rightarrow 3x+2y+z-14=0\end{array}$