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(D) Let the line be $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = t$. Any point $A$ on the line is given by $A(t, 2t+1, 3t+2)$.Since $A$ is the fo

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$11$

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(D) Let the line be $L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = t$. Any point $A$ on the line is given by $A(t, 2t+1, 3t+2)$.Since $A$ is the foot of the perpendicular from $Q(1, 6, 4)$ to the line,the vector $\overrightarrow{QA} = (t-1)\hat{i} + (2t-5)\hat{j} + (3t-2)\hat{k}$ must be perpendicular to the direction vector of the line $\vec{b} = \hat{i} + 2\hat{j} + 3\hat{k}$.Thus,$\overrightarrow{QA} \cdot \vec{b} = 0 \implies 1(t-1) + 2(2t-5) + 3(3t-2) = 0$.$t - 1 + 4t - 10 + 9t - 6 = 0 \implies 14t - 17 = 0 \implies t = \frac{17}{14}$.The foot of the perpendicular $A$ is $(\frac{17}{14}, 2(\frac{17}{14})+1, 3(\frac{17}{14})+2) = (\frac{17}{14}, \frac{48}{14}, \frac{79}{14})$.Since $A$ is the midpoint of $PQ$,where $P(\alpha, \beta, \gamma)$ is the image of $Q(1, 6, 4)$,we have $\frac{\alpha+1}{2} = \frac{17}{14}$,$\frac{\beta+6}{2} = \frac{48}{14}$,and $\frac{\gamma+4}{2} = \frac{79}{14}$.$\alpha = \frac{17}{7} - 1 = \frac{10}{7}$,$\beta = \frac{48}{7} - 6 = \frac{6}{7}$,$\gamma = \frac{79}{7} - 4 = \frac{51}{7}$.Then $2\alpha + \beta + \gamma = 2(\frac{10}{7}) + \frac{6}{7} + \frac{51}{7} = \frac{20+6+51}{7} = \frac{77}{7} = 11$.

Let $P(\alpha, \beta, \gamma)$ be the image of the point $Q(1, 6, 4)$ in the line $\frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3}$. Then $2\alpha + \beta + \gamma$ is equal to ..............

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