6. Let the circumradius of $\triangle A B C$ be $R, \frac{a \cos A+b \cos B+c \cos C}{a \sin B+b \sin C+c \sin A}=\frac{a+b+c}{9 R}$. Here $a, b, c$ are the sides opposite to $\angle A, \angle B, \angle C$ respectively. Try to find the sizes of the three interior angles of $\triangle A B C$.
6. As shown in the figure, let $O$ be the circumcenter of $\triangle ABC$. We have $9 R(a \cos A+b \cos B+c \cos C)=$ $18\left(S_{\triangle A B O}+S_{\triangle A C O}+S_{\triangle B C O}\right)=18 S_{\triangle A B C}$

On the other hand, by Chebyshev's inequality, we have $(a+b+c)(a \sin B+b \sin C+c \sin A) \geqslant$ $3(a b \sin C+b c \sin A+c a \sin B)=18 S_{\triangle A B C}$, equality holds if and only if $a: b: c=b \sin C$ : $c \sin A: a \sin B$. Since $a b \sin C=c a \sin B=b c \sin A$.
Therefore, $a=b=c$ and hence $A=B=C=60^{\circ}$