JEE Main 2025 Apr 8 Shift 2, Mathematics Q23: Lines in Space

Parametrise each line to make intersections easy to solve.

Line $L_1$: setting $x+2=y-1=z=t$ gives points $(t-2,t+1,t)$.

Line $L_2$: setting $\frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=s$ gives points $(5s+3,-s,s+1)$.

Line $L_3$: setting $\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}=u$ gives points $(-3u,3u+3,u+2)$.

Vertex $A=L_1\cap L_2$: equating coordinates gives $t=s+1$ from the third component; substituting into $t+1=-s$ yields $s=-1$, $t=0$, so $A=(-2,1,0)$.

Vertex $B=L_2\cap L_3$: equating gives $s=u+1$; substituting into $-s=3u+3$ yields $u=-1$, $s=0$, so $B=(3,0,1)$.

Vertex $C=L_3\cap L_1$: equating gives $u+2=t$ and $3u+3=t+1$; solving yields $u=0$, $t=2$, so $C=(0,3,2)$.

Form edge vectors from $A$: $\overrightarrow{AB}=(5,-1,1)$ and $\overrightarrow{AC}=(2,2,2)$.

Compute the cross product: $\overrightarrow{AB}\times \overrightarrow{AC}=\hat{i}(-2-2)-\hat{j}(10-2)+\hat{k}(10+2)=(-4,-8,12)$.

Magnitude: $|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{16+64+144}=\sqrt{224}=4\sqrt{14}$.

Area of the triangle: $A=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|=2\sqrt{14}$.

Therefore $A^2=4\cdot 14=56$.