JEE Main 2025 — 2 April, Evening Shift — Question 33

Let $X$ be mid point of $P$ and $Q$, which would be also feet of perpendicular.

Let $X$ divides $A$ and $B$ in $\lambda: 1, \lambda \neq-1$

$x=\left(\frac{3 \lambda+4}{\lambda+1}, \frac{5 \lambda+7}{\lambda+1}, \frac{3 \lambda+1}{\lambda+1}\right)$

Now $P X \perp A B \Rightarrow \overrightarrow{P X} \cdot \overrightarrow{A B}=0$

$\left(\frac{3 \lambda+4}{\lambda+1}-1\right) \cdot(4-3)+\left(\frac{5 \lambda+7}{\lambda+1}-0\right)(7-5)$

$$+\left(\frac{3 \lambda+1}{\lambda+1}-3\right) \cdot(1-3)=0$$

$\frac{2 \lambda+3}{\lambda+1}+\frac{10 \lambda+14}{\lambda+1}+\frac{4}{\lambda+1}=0$

$\Rightarrow \frac{12 \lambda+21}{\lambda+1}=0 \Rightarrow \lambda=\frac{-7}{4}$

$X=\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right)$

$X$ is mid point of $P Q$

$Q \equiv\left(2 \cdot \frac{5}{3}-1,2 \cdot \frac{7}{3}-0,2 \cdot \frac{17}{3}-3\right) \equiv(\alpha, \beta, \gamma)$

$\Rightarrow \alpha+\beta+\gamma=\frac{2(5+7+17)}{3}-4=\frac{58}{3}-4=\frac{46}{3}$