2.3.5 * In $\triangle A B C$, let $B C=a, C A=b, A B=c$. If $9 a^{2}+9 b^{2}-19 c^{2}=0$, then $\frac{\cot C}{\cot A+\cot B}=$ $\qquad$ .
Using the Law of Sines and Cosines, we have $\frac{\cot C}{\cot A+\cot B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}-\frac{\sin A \sin B}{\sin ^{2} C}$. $\cos C=\frac{a b}{c^{2}} \cdot \frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{a^{2}+b^{2}-c^{2}}{2 c^{2}}=\frac{9 a^{2}+9 b^{2}-9 c^{2}}{18 c^{2}}=\frac{19 c^{2}-9 c^{2}}{18 c^{2}}=\frac{5}{9}$.