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(C) Let the line be $L: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5} = \lambda$. Any point $M$ on the line is given by $M(2\lambda+1, 3\lambda-1, 5\l

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$14$

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(C) Let the line be $L: \frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5} = \lambda$. Any point $M$ on the line is given by $M(2\lambda+1, 3\lambda-1, 5\lambda+2)$.Since $M$ is the foot of the perpendicular from $A(8, 5, 7)$ to the line,the vector $\overrightarrow{AM}$ is perpendicular to the direction vector of the line $\vec{v} = 2\hat{i} + 3\hat{j} + 5\hat{k}$.$\overrightarrow{AM} = (2\lambda+1-8)\hat{i} + (3\lambda-1-5)\hat{j} + (5\lambda+2-7)\hat{k} = (2\lambda-7)\hat{i} + (3\lambda-6)\hat{j} + (5\lambda-5)\hat{k}$.Since $\overrightarrow{AM} \cdot \vec{v} = 0$,we have:$2(2\lambda-7) + 3(3\lambda-6) + 5(5\lambda-5) = 0$$4\lambda - 14 + 9\lambda - 18 + 25\lambda - 25 = 0$$38\lambda - 57 = 0 \implies \lambda = \frac{57}{38} = \frac{3}{2}$.Substituting $\lambda = \frac{3}{2}$ into $M$,we get $M(2(\frac{3}{2})+1, 3(\frac{3}{2})-1, 5(\frac{3}{2})+2) = M(4, \frac{7}{2}, \frac{19}{2})$.Let $A'(\alpha, \beta, \gamma)$ be the image of $A$. Since $M$ is the midpoint of $AA'$,we have:$\frac{\alpha+8}{2} = 4 \implies \alpha = 0$$\frac{\beta+5}{2} = \frac{7}{2} \implies \beta = 2$$\frac{\gamma+7}{2} = \frac{19}{2} \implies \gamma = 12$Thus,$\alpha + \beta + \gamma = 0 + 2 + 12 = 14$.

Let $(\alpha, \beta, \gamma)$ be the image of the point $A(8, 5, 7)$ in the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z-2}{5}$. Then $\alpha + \beta + \gamma$ is equal to

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