If the image of the point (4,4,3) in the line | vedclass

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

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$9$

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(A) Let the given line be $L: \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3} = \lambda$. Any point on the line is $Q(2\lambda+1, \lambda+2, 3\lambda+1)$. Let $P = (4,4,3)$. The vector $\vec{PQ} = (2\lambda+1-4, \lambda+2-4, 3\lambda+1-3) = (2\lambda-3, \lambda-2, 3\lambda-2)$. Since $\vec{PQ}$ is perpendicular to the line with direction vector $\vec{v} = (2, 1, 3)$,we have $\vec{PQ} \cdot \vec{v} = 0$. $2(2\lambda-3) + 1(\lambda-2) + 3(3\lambda-2) = 0$. $4\lambda - 6 + \lambda - 2 + 9\lambda - 6 = 0$. $14\lambda - 14 = 0 \Rightarrow \lambda = 1$. Thus,the foot of the perpendicular $Q$ is $(2(1)+1, 1+2, 3(1)+1) = (3, 3, 4)$. Let the image of $P$ be $R(\alpha, \beta, \gamma)$. Since $Q$ is the midpoint of $PR$,we have: $\frac{\alpha+4}{2} = 3 \Rightarrow \alpha = 2$. $\frac{\beta+4}{2} = 3 \Rightarrow \beta = 2$. $\frac{\gamma+3}{2} = 4 \Rightarrow \gamma = 5$. So,the image is $(2, 2, 5)$. Therefore,$\alpha+\beta+\gamma = 2+2+5 = 9$.

If the image of the point $(4,4,3)$ in the line $\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-1}{3}$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

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