JEE Main 2025 Jan 22 Shift 1, Mathematics Q24: Lines in Space

Parametrise: a point on $L_1$ is $(3\lambda +1,-\lambda +1,-1)$; a point on $L_2$ is $(2\mu +2,0,\alpha \mu -4)$. At intersection B these match:

$3\lambda +1=2\mu +2$, $-\lambda +1=0\Rightarrow \lambda =1$, so $3+1=2\mu +2\Rightarrow \mu =1$. Then $-1=\alpha \cdot 1-4\Rightarrow \alpha =3$.

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

Foot of perpendicular from $A=(1,1,-1)$ on $L_2$: parameterise $P=(2\delta +2,0,3\delta -4)$. Then $\overrightarrow{AP}=(2\delta +1,-1,3\delta -3)$. Setting $\overrightarrow{AP}\cdot (2,0,3)=0$:

$2(2\delta +1)+3(3\delta -3)=0\Rightarrow 4\delta +2+9\delta -9=0\Rightarrow 13\delta =7\Rightarrow \delta =\frac{7}{13}$.

$P=\left(\frac{14}{13}+2,0,\frac{21}{13}-4\right)=\left(\frac{40}{13},0,-\frac{31}{13}\right)$.

$(PB{)}^2={\left(4-\frac{40}{13}\right)}^2+0+{\left(-1+\frac{31}{13}\right)}^2={\left(\frac{12}{13}\right)}^2+{\left(\frac{18}{13}\right)}^2=\frac{144+324}{169}=\frac{468}{169}$.

$26\alpha (PB{)}^2=26\cdot 3\cdot \frac{468}{169}=\frac{78\cdot 468}{169}=\frac{36504}{169}=216$.