Let (alpha, beta, gamma) be the mirror image | vedclass

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(B) Let $P = (2, 3, 5)$ and $R = (\alpha, \beta, \gamma)$ be its mirror image in the line $L: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$.Let $M$ b

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$33$

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(B) Let $P = (2, 3, 5)$ and $R = (\alpha, \beta, \gamma)$ be its mirror image in the line $L: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$.Let $M$ be the midpoint of $PR$. Since $R$ is the mirror image of $P$ in the line $L$,the line segment $PR$ is perpendicular to the line $L$.The direction vector of the line $L$ is $\vec{v} = (2, 3, 4)$.The vector $\vec{PR} = (\alpha - 2, \beta - 3, \gamma - 5)$.Since $\vec{PR} \perp \vec{v}$,their dot product must be zero:$\vec{PR} \cdot \vec{v} = 0$$(\alpha - 2, \beta - 3, \gamma - 5) \cdot (2, 3, 4) = 0$$2(\alpha - 2) + 3(\beta - 3) + 4(\gamma - 5) = 0$$2\alpha - 4 + 3\beta - 9 + 4\gamma - 20 = 0$$2\alpha + 3\beta + 4\gamma = 4 + 9 + 20$$2\alpha + 3\beta + 4\gamma = 33$.

Let $(\alpha, \beta, \gamma)$ be the mirror image of the point $(2, 3, 5)$ in the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$. Then $2\alpha + 3\beta + 4\gamma$ is equal to

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