Let a plane $P$ contain two lines
$\vec{r}=\hat{i}+\lambda(\hat{i}+\hat{j}), \lambda \in \mathbf{R}$ and $\vec{r}=-\hat{j}+\mu(\hat{j}-\hat{k}), \mu \in \mathbf{R}$.
If $Q(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from
the point $M(1,0,1)$ to $P$, then $3(\alpha+\beta+\gamma)$ equals__________.
Normal of plane $=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 0 & 1 & -1\end{array}\right|$
$\vec{n}=-\hat{i}+\hat{j}+\hat{k}$
Direction ratios of normal to the plane $=\langle-1,1,1\rangle$
Equation of plane
$-1(x-1)+1(y-0)+1(z-0)=0$
$\Rightarrow x-y-z-1=0$
If $(x, y, z)$ is foot of perpendicular of $M(1,0,1)$ on the plane then
$\Rightarrow \frac{x-1}{1}=\frac{y-0}{-1}=\frac{z-1}{-1}=\frac{-(1-0-1-1)}{3}$
$\therefore x=\frac{4}{3}, y=-\frac{1}{3}, z=\frac{2}{3}$
$\alpha+\beta+\gamma=\frac{4}{3}-\frac{1}{3}+\frac{2}{3}=\frac{5}{3}$
$\therefore 3(\alpha+\beta+\gamma)=3 \times \frac{5}{3}=5$