Let a plane P contain two lines overrightarro | vedclass

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(C) The direction vectors of the two lines are $\vec{v}_1 = \hat{i} + \hat{j}$ and $\vec{v}_2 = \hat{j} - \hat{k}$.The normal vector $\vec{n}$ to the

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$5$

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(C) The direction vectors of the two lines are $\vec{v}_1 = \hat{i} + \hat{j}$ and $\vec{v}_2 = \hat{j} - \hat{k}$.The normal vector $\vec{n}$ to the plane $P$ is given by the cross product $\vec{v}_1 	imes \vec{v}_2$:$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 0 \ 0 & 1 & -1 \end{vmatrix} = \hat{i}(-1) - \hat{j}(-1) + \hat{k}(1) = -\hat{i} + \hat{j} + \hat{k}$.The plane passes through the point $(1, 0, 0)$ (from the first line). Thus,the equation of the plane is:$-1(x - 1) + 1(y - 0) + 1(z - 0) = 0 \implies -x + y + z + 1 = 0 \implies x - y - z - 1 = 0$.Let $Q(\alpha, \beta, \gamma)$ be the foot of the perpendicular from $M(1, 0, 1)$ to the plane $x - y - z - 1 = 0$. The line passing through $M$ and perpendicular to the plane is:$\frac{\alpha - 1}{1} = \frac{\beta - 0}{-1} = \frac{\gamma - 1}{-1} = k$.So,$\alpha = k + 1, \beta = -k, \gamma = 1 - k$.Since $Q$ lies on the plane:$(k + 1) - (-k) - (1 - k) - 1 = 0 \implies k + 1 + k - 1 + k - 1 = 0 \implies 3k - 1 = 0 \implies k = \frac{1}{3}$.Thus,$\alpha = \frac{1}{3} + 1 = \frac{4}{3}, \beta = -\frac{1}{3}, \gamma = 1 - \frac{1}{3} = \frac{2}{3}$.Finally,$3(\alpha + \beta + \gamma) = 3(\frac{4}{3} - \frac{1}{3} + \frac{2}{3}) = 3(\frac{5}{3}) = 5$.

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