6. In $\triangle A B C$, the sides opposite to angles $A$, $B$, and $C$ are $a$, $b$, and $c$ respectively. Given $a^{2}+b^{2}-7 c^{2}=0$, then $\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=$
6.3 .
$$
\begin{array}{l}
\frac{\operatorname{ctg} C}{\operatorname{ctg} A+\operatorname{ctg} B}=\frac{\frac{\cos C}{\sin C}}{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}}=\frac{\cos C \sin A \sin B}{\sin (A+B) \sin C} \\
=\cos C \cdot \frac{\sin A \sin B}{\sin ^{2} C}=\frac{a^{2}+b^{2}-c^{2}}{2 a b} \cdot \frac{a b}{c^{2}} \\
=\frac{7 c^{2}-c^{2}}{2 c^{2}}=3 .
\end{array}
$$