JEE Main 2025 — 4 April, Morning Shift — Question 37

$L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}$ Point $A(3 \lambda+6,2 \lambda+7,7-2 \lambda)$

$B(3 \mu+6,2 \mu+7,7-2 \mu)$

Let $P(3 k+6,2 k+7,7-2 k)$ be foot of perpendicular from $P^{\prime}(1,2,3)$

$\therefore P P^{\prime}<3,2,-2>=0$

$3(3 k+5)+(2 k+5) 2+(4-2 k)(-2)=0$

$9 k+15+4 k+10-8+4 k=0$ $17 k+17=0$

$\Rightarrow k=-1$

$\therefore \quad P(3,5,9)$

$|\overrightarrow{A P}|=2 \sqrt{17}$ }

$\Rightarrow(3 \lambda+3)^{2}+(2 \lambda+2)^{2}+(-2-2 \lambda)^{2}=17 \times 4$ $=17(\lambda+1)^{2}=17 \times 4$

$\Rightarrow \lambda+1= \pm 2 \Rightarrow \lambda=1$ or $\lambda=-3$

$\therefore \quad A(9,9,5), B(-3,1,13)$

$\overrightarrow{O A} \cdot \overrightarrow{O B}=(9 \hat{i}+9 \hat{j}+5 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+13 \hat{k})$

$=-27+9+65=47$