If $\overline{\mathrm{a}} , \overline{\mathrm{b}}$  are such that  $\left|\overline{\mathrm{a}}\right|=\sqrt{3} , \left|\overline{\mathrm{b}}\right|=2$   and $\left(\overline{\mathrm{a}},\overline{\mathrm{b}}\right)=\frac{\mathrm{\pi }}{3}$ . Then the area of the triangle with adjacent sides   $\overline{\mathrm{a}}+2\overline{\mathrm{b}}$ and $2\overline{\mathrm{a}}+\overline{\mathrm{b}}$  is

$\begin{array}{l}\left|\overline{\mathrm{a}}\right|=\sqrt{3} , \left|\overline{\mathrm{b}}\right|=2\\ \text{ Area of triangle is }=\frac{1}{2}\left|\right(\overline{\mathrm{a}}+2\overline{\mathrm{b}})\times (2\overline{\mathrm{a}}+\overline{\mathrm{b}}\left)\right|\\ =\frac{1}{2}\left|\right(\overline{\mathrm{a}}\times \overline{\mathrm{b}})+4(\overline{\mathrm{b}}\times \overline{\mathrm{a}}\left)\right|\Rightarrow \frac{1}{2}|-(\overline{\mathrm{b}}\times \overline{\mathrm{a}})+4(\overline{\mathrm{b}}\times \overline{\mathrm{a}}\left)\right|\Rightarrow \frac{1}{2}\left|3\right(\overline{\mathrm{b}}\times \overline{\mathrm{a}}\left)\right|\\ =\frac{3}{2}\left|\right(\overline{\mathrm{b}}\times \overline{\mathrm{a}}\left)\right|=\frac{3}{2}\left|\overline{\mathrm{b}}\right|\left|\overline{\mathrm{a}}\right|\sin \left(\overline{a},\overline{b}\right)\Rightarrow \frac{3}{2}\left(\sqrt{3}\right)\left(2\right)\sin \frac{\mathrm{\pi }}{3}\\ =\frac{3}{2}\left(\sqrt{3}\right)\left(2\right)\frac{\sqrt{3}}{2}=\frac{9}{2}\end{array}$