Question
If $\left( 2\alpha + 1,\alpha^{2} - 3\alpha,\frac{\alpha - 1}{2} \right)$ is the image of $(\alpha,2\alpha,1)$ in the line $\frac{x - 2}{3} = \frac{y - 1}{2} = \frac{z}{1}$, then the possible value(s) of $\alpha$ is(are) :
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Solution
Sol. [Key Concept: The point is the mirror image, so it is the foot's reflection; perpendicularity plus midpoint-on-line fixes $\alpha = 3$.]
$\ \therefore P$ is mid point of AB and lies on the given line
$$\Rightarrow P\left( \frac{3\alpha + 1}{2},\frac{\alpha^{2} - \alpha}{2},\frac{\alpha + 1}{4} \right) $$
Now put coordinates of P in the line
$${\Rightarrow \frac{\frac{3\alpha + 1}{2} - 2}{3} = \frac{\frac{\alpha^{2} - \alpha}{2} - 1}{2} = \frac{\alpha + 1}{4} }{\Rightarrow \frac{\alpha - 1}{2} = \frac{\alpha + 1}{4} = \frac{\alpha^{2} - \alpha - 2}{4} }{\Rightarrow \alpha = 3 }$$Now for $\alpha = 3$, clearly AB is perpendicular to $3\widehat{i} + 2\widehat{j} + k$
for $\alpha = 3$ point B is image of point A
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