If the image of the point P(1, 0, 3) in the l | vedclass

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(B) The line passing through $A(4, 7, 1)$ and $B(3, 5, 3)$ has direction ratios $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation of the line $AB$ is $\fra

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$\frac{46}{3}$

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(B) The line passing through $A(4, 7, 1)$ and $B(3, 5, 3)$ has direction ratios $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation of the line $AB$ is $\frac{x-3}{-1} = \frac{y-5}{-2} = \frac{z-3}{2} = \lambda$.Any point $R$ on the line is $(\lambda+3, 2\lambda+5, -2\lambda+3)$.The vector $\vec{PR} = (\lambda+3-1, 2\lambda+5-0, -2\lambda+3-3) = (\lambda+2, 2\lambda+5, -2\lambda)$.Since $PR \perp AB$,the dot product of $\vec{PR}$ and the direction vector of $AB$ $(-1, -2, 2)$ is $0$:$-1(\lambda+2) - 2(2\lambda+5) + 2(-2\lambda) = 0$.$-\lambda - 2 - 4\lambda - 10 - 4\lambda = 0 \Rightarrow -9\lambda = 12 \Rightarrow \lambda = -\frac{4}{3}$.The foot of the perpendicular $R$ is $(\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$.Since $R$ is the midpoint of $PQ$,$Q = 2R - P = (2 	imes \frac{5}{3} - 1, 2 	imes \frac{7}{3} - 0, 2 	imes \frac{17}{3} - 3) = (\frac{7}{3}, \frac{14}{3}, \frac{25}{3})$.Thus,$\alpha + \beta + \gamma = \frac{7+14+25}{3} = \frac{46}{3}$.

If the image of the point $P(1, 0, 3)$ in the line joining the points $A(4, 7, 1)$ and $B(3, 5, 3)$ is $Q(\alpha, \beta, \gamma)$,then $\alpha + \beta + \gamma$ is equal to

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