JEE 2025 — Mathematics PYQ

$$L_1:\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda \Rightarrow P(3\lambda +1,-\lambda +1,-1)$$

$$L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha }=\mu \Rightarrow Q(2\mu +2,0,\alpha \mu -4)$$

$$\because L_1\text{ and }{L}_2\text{ intersect}$$

$$\therefore 3\lambda +1=2\mu +2\mid -\lambda +1=0\mid -1=\alpha \mu -4$$

$$\lambda =1\Rightarrow 4=2\mu +2\Rightarrow \mu =1$$

$$-1=\alpha -4\Rightarrow \alpha =3$$

$$B=(4,0,-1)$$

$$\therefore L_2:\frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{3}$$

$$\overrightarrow{AM}\perp (2\hat{i}+0\hat{j}+3\hat{k})\Rightarrow \overrightarrow{AM}\cdot (2\hat{i}+3\hat{k})=0$$

$$\Rightarrow 2(2\mu +1)+3(3\mu -3)=0\Rightarrow 13\mu =7\Rightarrow \mu =\frac{7}{13}$$

$$\therefore 26\alpha (PB{)}^2=26(3)\frac{36}{13}=6\times 36=216$$