Let a plane P contain two lines
$\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}), \lambda \in \mathrm{R}$ and
$\overrightarrow{\mathrm{r}}=-\hat{\mathrm{j}}+\mu(\hat{\mathrm{j}}-\hat{\mathrm{k}}), \mu \in \mathrm{R}$
If $\mathrm{Q}(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from the point $M(1,0,1)$ to $P$, then $3(\alpha+\beta+\gamma)$ equals_______________
Dr's normal to plane
$=\left|\begin{array}{ccc}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 1 & 1 & 0 \\ 0 & 1 & -1\end{array}\right|=-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$
Equation of plane
$-1(x-1)+1(y-0)+1(z-0)=0$
$x-y-z-1=0$..........(1)
Now $\frac{\alpha-1}{1}=\frac{\beta-0}{-1}=\frac{\gamma-1}{-1}=-\frac{(1-0-1-1)}{3}$
$\frac{\alpha-1}{1}=\frac{\beta}{-1}=\frac{\gamma-1}{-1}=\frac{1}{3}$
$\alpha=\frac{4}{3}, \beta=-\frac{1}{3}, \gamma=\frac{2}{3}$
$3(\alpha+\beta+\gamma)=3\left(\frac{4}{3}-\frac{1}{3}+\frac{2}{3}\right)=5$
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