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JEE Main 2024
04-04-2024
Question
Consider a line L passing through the points $\mathrm{P}(1,2,1)$ and $\mathrm{Q}(2,1,-1)$. If the mirror image of the point $\mathrm{A}(2,2,2)$ in the line L is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+6 \gamma$ is equal to $\ldots .$.
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Solution

DR's of Line $L \equiv-1: 1: 2$
DR's of $A B \equiv \alpha-2: \beta-2: \gamma-2$
$ \begin{aligned} & \mathrm{AB} \perp_{\mathrm{ar}} \mathrm{~L} \Rightarrow 2-\alpha+\beta-2+2 \gamma-4=0 \\ & 2 \gamma+\beta-\alpha=4 \end{aligned} $
Let C is mid-point of AB
$ \mathrm{C}\left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right) $
DR's of $\mathrm{PC}=\frac{\alpha}{2}: \frac{\beta-2}{2}: \frac{\gamma}{2}$
line $\mathrm{L} \| \mathrm{PC} \Rightarrow \frac{-\alpha}{2}=\frac{\beta-2}{2}=\frac{\gamma}{4}=\mathrm{K}$ (let)
$ \begin{aligned} & \alpha=-2 \mathrm{~K} \\ & \beta=2 \mathrm{~K}+2 \\ & \gamma=4 \mathrm{~K} \end{aligned} $
use in (1) ⇒ K = $\frac{1}{6}$
value of $\alpha+\beta+6 \gamma=24 K+2=6$
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