JEE Main 2026 — 4 April, Evening Shift — Question 38

$\therefore \mathrm{P}$ is mid point of AB and lies on the given line $\Rightarrow \mathrm{P}\left(\frac{3 \alpha+1}{2}, \frac{\alpha^{2}-\alpha}{2}, \frac{\alpha+1}{4}\right)$ 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 $3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\mathrm{k}$ for $\alpha=3$ point B is image of point A