Consider a line $\mathrm{L}$ 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 $\mathrm{L}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+6 \gamma$ is equal to _______

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Aicharya · Accepted Answer

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

Consider a line $\mathrm{L}$ passing through the points $\mathrm{P}(1,2,1)$ and $\mathrm{Q}(2,1,-1)$. If the…

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