Consider a line L passing through the points | vedclass

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(A) The direction ratios of line $L$ passing through $P(1, 2, 1)$ and $Q(2, 1, -1)$ are $(2-1, 1-2, -1-1) = (1, -1, -2)$.The equation of line $L$ is $

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$6$

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(A) The direction ratios of line $L$ passing through $P(1, 2, 1)$ and $Q(2, 1, -1)$ are $(2-1, 1-2, -1-1) = (1, -1, -2)$.The equation of line $L$ is $\frac{x-1}{1} = \frac{y-2}{-1} = \frac{z-1}{-2} = k$.Any point $C$ on line $L$ is $(k+1, -k+2, -2k+1)$.Since $AC$ is perpendicular to line $L$,the direction ratios of $AC$ are $(k+1-2, -k+2-2, -2k+1-2) = (k-1, -k, -2k-1)$.Since $AC \perp L$,the dot product of their direction ratios is zero:$1(k-1) + (-1)(-k) + (-2)(-2k-1) = 0$$k - 1 + k + 4k + 2 = 0$$6k + 1 = 0 \Rightarrow k = -\frac{1}{6}$.Substituting $k$ in the coordinates of $C$,we get $C = (1 - \frac{1}{6}, 2 + \frac{1}{6}, 1 + \frac{2}{6}) = (\frac{5}{6}, \frac{13}{6}, \frac{8}{6})$.Since $C$ is the midpoint of $AB$,where $B = (\alpha, \beta, \gamma)$ and $A = (2, 2, 2)$:$\frac{\alpha+2}{2} = \frac{5}{6} \Rightarrow \alpha+2 = \frac{5}{3} \Rightarrow \alpha = -\frac{1}{3}$.$\frac{\beta+2}{2} = \frac{13}{6} \Rightarrow \beta+2 = \frac{13}{3} \Rightarrow \beta = \frac{7}{3}$.$\frac{\gamma+2}{2} = \frac{8}{6} \Rightarrow \gamma+2 = \frac{8}{3} \Rightarrow \gamma = \frac{2}{3}$.Now,$\alpha + \beta + 6\gamma = -\frac{1}{3} + \frac{7}{3} + 6(\frac{2}{3}) = \frac{6}{3} + 4 = 2 + 4 = 6$.

Consider a line $L$ passing through the points $P(1, 2, 1)$ and $Q(2, 1, -1)$. If the mirror image of the point $A(2, 2, 2)$ in the line $L$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + 6\gamma$ is equal to:

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