JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Straight Lines

Let the equilateral triangle have side $a$ and let $\hat{n}$ be the direction perpendicular to the two lines. Since $B$ lies on the line at distance $6$ and $C$ on the line at distance $3$ on the opposite side, the components of the sides along $\hat{n}$ give $a\cos \alpha =6$ and $a\cos (\alpha +60^{\circ })=-3$, where $\alpha$ is the angle $AB$ makes with $\hat{n}$. Expanding the second, $3-\frac{\sqrt{3}}{2}a\sin \alpha =-3$, so $a\sin \alpha =4\sqrt{3}$. Hence $a^2=(a\cos \alpha {)}^2+(a\sin \alpha {)}^2=36+48=84$, and the area is $\frac{\sqrt{3}}{4}{a}^2=21\sqrt{3}$.