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Let a plane P contains two lines \vec{r}=\hat{i}+\lambda (\hat{i}+\hat{j}),\lambda \; \epsilon \; R and\vec{r}=-\hat{j}+\mu (\hat{j}-\hat{k}),\mu \; \epsilon \; R. If Q\left ( \alpha ,\beta ,\gamma \right ) is the foot of the perpendicular drawn from the point M(1,0,1) to P, then 3(\alpha +\beta +\gamma ) equals ______.
Option: 1 10
Option: 2 6
Option: 3 5
Option: 4 8

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best_answer

Sol. Dr’s normal to plane

=\left|\begin{array}{ccc} i & j & k \\ 1 & 1 & 0 \\ 0 & 1 & -1 \end{array}\right|=-\hat{i}+\hat{j}+\hat{k}

Equation of plane

\begin{aligned} &-1(x-1)+1(y-0)+1(z-0)=0\\ &x-y-z-1=0\;\;\;\;\;\;\;\;\ldots(1) \end{aligned}

\\\text { 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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