Question:
A plane passes through the points $\mathrm{A}(1,2,3), \mathrm{B}(2,3,1)$ and $\mathrm{C}(2,4,2)$. If $\mathrm{O}$ is the origin and $\mathrm{P}$ is $(2,-1,1)$,
Correct Option: , 3
Solution:
Normal to plane $\overrightarrow{\mathrm{n}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 1 & -2 \\ 0 & 1 & 1\end{array}\right|$
$=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\overrightarrow{\mathrm{OP}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$
$\cos \theta=\frac{6+1+1}{\sqrt{6} \sqrt{11}}=\frac{8}{\sqrt{66}} \Rightarrow \sin \theta=\sqrt{\frac{2}{66}}$
$\therefore$ Projection of $\overrightarrow{\mathrm{OP}}$ on plane $=|\overrightarrow{\mathrm{OP}}| \sin \theta$
$=\sqrt{\frac{2}{11}}$
option (3)