JEE Main 2026 — 6 April, Morning Shift — Question 35

Let Q be the point $\perp^{\mathrm{r}}$ from $\mathrm{P}(1,6, \mathrm{a})$ to the line $\mathrm{L}: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-\mathrm{a}+1}{\mathrm{~b}}$ The image of point P in line L is $\mathrm{P}^{\prime}\left(\frac{\mathrm{a}}{3}, 0, \mathrm{a}+\mathrm{c}\right)$ ∵ Q is the mid point $\mathrm{PP}^{\prime}$ : $\mathrm{Q} \equiv\left(\frac{1+\frac{\mathrm{a}}{3}}{2}, \frac{6+0}{2}, \frac{\mathrm{a}+\mathrm{a}+\mathrm{c}}{2}\right) \equiv\left(\frac{3+\mathrm{a}}{6}, 3, \frac{2 \mathrm{a}+\mathrm{c}}{2}\right)$ comparing $x$ and $y, \frac{3+a}{6}=1 \Rightarrow a=3$ $\therefore Q \equiv\left(1,3, \frac{6+c}{2}\right)$ Direction vector of L is $\overrightarrow{\mathrm{v}}=(1,2, \mathrm{~b})$ vector PQ is $\perp^{\mathrm{r}}$ to $\overrightarrow{\mathrm{V}}$ $\therefore \overrightarrow{\mathrm{PQ}}=\left(1-1,3-6, \frac{6+\mathrm{c}}{2}-3\right)$ $=\left(0,-3, \frac{\mathrm{c}}{2}\right)$ $\overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{v}}=0$ $(0) .(1)+(-3)(2)+\left(\frac{\mathrm{c}}{2}\right)(\mathrm{b})=0$ $\mathrm{bc}=12$ Now comparing z with $\mathrm{Q}\left(1,3, \frac{6+\mathrm{c}}{2}\right)$ $\frac{3-1}{2}=\frac{\frac{6+c}{2}-3+1}{b}$ $2 \mathrm{~b}-\mathrm{c}=2$ solving $b c=12$ and $c=2 b-2$ $\mathrm{b}=3,-2$ $\mathrm{b}=3 \quad(\mathrm{~b}>0)$ $\mathrm{c}=4$ $\therefore \mathrm{Q}(1,3,5)$ Now, any point S on the line L can be written as $(\mathrm{k}, 2 \mathrm{k}+1,3 \mathrm{k}+2)$ $S Q=\sqrt{(k-1)^{2}+(2 k-2)^{2}+(3 k-3)^{2}}=2 \sqrt{14}$ $\mathrm{k}=3,-1$ for $\mathrm{k}=3, \mathrm{~S} \equiv(3,7,11)$ $\therefore \alpha+\beta+\gamma=21$