2. Given $\triangle A B C$ with angles $A, B, C$ opposite to sides $a, b, c$ respectively, and $a^{2}+b^{2}=c^{2}+\frac{2}{3} a b$. If the circumradius of $\triangle A B C$ is $\frac{3 \sqrt{2}}{2}$, then the maximum area of $\triangle A B C$ is $\qquad$.
2. $4 \sqrt{2}$ Detailed Explanation: $\because a^{2}+b^{2}=c^{2}+\frac{2}{3} a b, \therefore \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{1}{3}$,
$\therefore \cos C=\frac{1}{3}$, then $\sin C=\frac{2 \sqrt{2}}{3}$, and since $c=2 R \sin C=2 \times \frac{3}{2} \sqrt{2} \times \frac{2 \sqrt{2}}{3}=4$,
$\therefore c^{2}=a^{2}+b^{2}-2 a b \cos C \Rightarrow 16+\frac{2}{3} a b=a^{2}+b^{2} \geqslant 2 a b$,
$\therefore a b \leqslant 12 . \because S=\frac{1}{2} a b \sin C=\frac{\sqrt{2}}{3} a b \leqslant 4 \sqrt{2}$,
$\therefore$ when $a=b=2 \sqrt{3}$, the area of triangle $ABC$ has a maximum value of $4 \sqrt{2}$.