Question

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

Answer 1

Let $\overrightarrow{n}$ be the normal vector of the given plane
$\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 2& 0\\ 1& -1& 3\end{array}\right|=6\hat{i}-2\hat{k}=2(3\hat{i}-\hat{k})$

$\because \overrightarrow{a}$ is perpendicular to $\overrightarrow{n}$ and $\hat{i}+2\hat{j}+3\hat{k},$ $\therefore \overrightarrow{a}=\lambda \left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& -1\\ 1& 2& 3\end{array}\right|=\lambda (2\hat{i}-10\hat{j}+6\hat{k})$

$\because \overrightarrow{a}\cdot (i+j+2k)=2\therefore \lambda (2-10+12)=2\Rightarrow \lambda =\frac{1}{2}$
And $\overrightarrow{a}=\hat{i}-5\hat{j}+3\hat{k}$
$\Rightarrow \alpha =1,\beta =-5,\gamma =3$
Now, $(\alpha -\beta +\gamma {)}^2=(1+5+3{)}^2=81$