JEE Main · 2023 · 01 Feb 2023 · Shift 1 · Properties of Triangle

Minimizing $\cos 2A+\cos 2B+\cos 2C$ forces $A=B=C=60^{\circ }$, so the triangle is equilateral. With inradius $r=3$, the side is $a=2\sqrt{3}r=6\sqrt{3}$, giving perimeter $18\sqrt{3}$ and area $\frac{\sqrt{3}}{4}(6\sqrt{3}{)}^2=27\sqrt{3}$. Each vertex lies at distance $6$ from the incentre with $120^{\circ }$ between them, so $\overrightarrow{MA}\cdot \overrightarrow{MB}=36\cos 120^{\circ }=-18$, while $\sin 2A+\sin 2B+\sin 2C=\sin A+\sin B+\sin C=\frac{3\sqrt{3}}{2}$. The statement that the area is $\frac{27\sqrt{3}}{2}$ is the one that does not hold.