JEE Main 2026 — 21 January, Morning Shift — Question 4

Let the distance between L1 and L2 be $d = 6 + 3 = 9$. Let side length of equilateral triangle ABC be $a$. Let $\theta$ be the angle between line AB and the perpendicular to L1. Then distance from A to L1: $a \sin \theta = 6$. Distance from A to L2: $a \sin(60^\circ - \theta) = 3$. Using $\sin(60^\circ - \theta) = \sin 60^\circ \cos \theta - \cos 60^\circ \sin \theta = \frac{\sqrt{3}}{2} \cos \theta - \frac{1}{2} \sin \theta$. Substitute $\sin \theta = \frac{6}{a}$ and $\cos \theta = \sqrt{1 - \frac{36}{a^2}}$: $a\left(\frac{\sqrt{3}}{2} \sqrt{1 - \frac{36}{a^2}} - \frac{1}{2} \cdot \frac{6}{a}\right) = 3$. Simplify: $\frac{\sqrt{3}}{2} \sqrt{a^2 - 36} - 3 = 3$ → $\frac{\sqrt{3}}{2} \sqrt{a^2 - 36} = 6$ → $\sqrt{a^2 - 36} = 4\sqrt{3}$. Square: $a^2 - 36 = 48$ → $a^2 = 84$. Area of equilateral triangle: $\frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \times 84 = 21\sqrt{3}$.