If $\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}$ is minimum then $\mathrm{A}=$ $\mathrm{B}=\mathrm{C}=60^{\circ}$ So $\triangle \mathrm{ABC}$ is equilateral Now in-radias $r=3$ So in $\triangle \mathrm{MBD}$ we have $\begin{aligned} & \operatorname{tan} 30^{\circ}=\frac{M D}{B D}=\frac{r}{a / 2}=\frac{6}{a} \\\\ & 1 / \sqrt{3}=\frac{1}{a}=a=6 \sqrt{3} \end{aligned}$ Perimeter of $\triangle \mathrm{ABC}=18 \sqrt{3}$ Area of $\triangle \mathrm{ABC}=\frac{\sqrt{3}}{4} a^2=27 \sqrt{3}$