Let a plane $P$ contain two lines $\overrightarrow{r}=\hat{i}+\lambda \left(\hat{i}+\hat{j}\right), \lambda \in R$ and $\overrightarrow{\mathrm{r}}=-\hat{\mathrm{j}}+\mu \left(\hat{\mathrm{j}}-\hat{\mathrm{k}}\right), \mu \in \mathrm{R}$. If $\mathrm{Q}\left(\alpha , \beta , \gamma \right)$ is the foot of the perpendicular drawn from the point $\mathrm{M}\left(1, 0, 1\right)$ to $\mathrm{P}$, then $3\left(\alpha +\beta +\gamma \right)$ equals .......

Normal vector to the plane containing to given lines $\overrightarrow{n}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 0\\ 0& 1& -1\end{array}\right|=-\hat{i}+\hat{j}+\hat{k}$

Direction ratios of normal vector $=-1, 1, 1$.

Equation of plane will be $-1\left(x-1\right)+1\left(y-0\right)+1\left(z-0\right)=0$

$\Rightarrow x-y-z-1=0$

If $\left(x, y, z\right)$ is foot of perpendicular drawn from $M\left(1, 0, 1\right)$ on the plane, then

$\Rightarrow \frac{x-1}{1}=\frac{y-0}{-1}=\frac{z-1}{-1}=\frac{-\left(1-0-1-1\right)}{3}$

$\Rightarrow x=\frac{4}{3}, y=-\frac{1}{3}, z=\frac{2}{3}$

$\Rightarrow \alpha +\beta +\gamma =\frac{4}{3}-\frac{1}{3}+\frac{2}{3}=\frac{5}{3}$

$\Rightarrow 3\left(\alpha +\beta +\gamma \right)=5$.