JEE Main · 2025 · 04 Apr 2025 · Shift 2 · Three Dimensional Geometry

The lines intersect where their coordinates agree. On $L_1$ the $y$-coordinate is fixed at $5$, and setting $-3+4s=5$ on $L_2$ gives $s=2$, returning the common point $A(7,5,3)$. The triangle is isosceles with $AB=AC=\sqrt{15}$, so its area is $\frac{1}{2}AB\cdot AC\sin \theta$, where $\theta$ is the angle between the lines. With direction vectors $(1,0,-1)$ and $(3,4,5)$, $\cos \theta =\frac{|3-5|}{\sqrt{2}\sqrt{50}}=\frac{1}{5}$, hence $\sin \theta =\frac{2\sqrt{6}}{5}$. The area is $\frac{1}{2}\cdot 15\cdot \frac{2\sqrt{6}}{5}=3\sqrt{6}$, and the square of the area is $(3\sqrt{6}{)}^2=54$.