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Let $(\alpha, \beta, \gamma)$ be the image of the point $(8,5,7)$ 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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$\begin{aligned} & \overrightarrow{\mathrm{AM}} \cdot(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})=0 \\ & (2 \lambda-7)(2)+(3 \lambda-6)(3)+(5 \lambda-5)(5)=0 \\ & 38 \lambda=57 \\ & \lambda=\frac{3}{2} \\ & \mathrm{M}\left(4, \frac{7}{2}, \frac{19}{2}\right) \\ & \mathrm{A}^{\prime}(0,2,12)\end{aligned}$
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