# Let P be a plane passing

Question:

Let $P$ be a plane passing through the points $(1,0,1),(1,-2,1)$ and $(0,1,-2)$. Let a vector $\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}}$ be such that $\overrightarrow{\mathrm{a}}$ is parallel to the plane $P$, perpendicular to $(\hat{i}+2 \hat{j}+3 \hat{k})$ and $\overrightarrow{\mathrm{a}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=2$, then $(\alpha-\beta+\gamma)^{2}$ equals

Solution:

Equation of plane :

$\left|\begin{array}{ccc}x-1 & y-0 & z-1 \\ 1-1 & 2 & 1-1 \\ 1-0 & 0-1 & 1+2\end{array}\right|=0$

$\Rightarrow 3 x-z-2=0$

$\overrightarrow{\mathrm{a}}=\alpha \hat{\mathrm{i}}+\beta \hat{\mathrm{j}}+\gamma \hat{\mathrm{k}} \|$ to $3 \mathrm{x}-\mathrm{z}-2=0$

$\Rightarrow 3 \alpha-8=0$..(1)

$\vec{a} \perp i+2 \hat{j}+3 \hat{k}$

$\Rightarrow \alpha+2 \beta+38=0 \quad \ldots . .(2)$

$\overrightarrow{\mathrm{a}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}})=0$

$\Rightarrow \alpha+\beta+28=2$..(3)

on solving $1,2 \& 3$

$\alpha=1, \quad \beta=-5,8=3$

So $(\alpha-\beta+8)=81$