6. In $\triangle A B C$, $a, b, c$ are the sides opposite to the internal angles $A, B, C$ respectively. If $7 b^{2}+25 c^{2}-25 a^{2}=0$. Try to find the value of $\sin (2 A-C)$ when $A-C$ reaches its maximum value.
$$
\begin{array}{l}
7 b^{2}+25 c^{2}=25 a^{2}=25\left(b^{2}+c^{2}-2 b c \cos A\right) \Rightarrow 9 b^{2}-25 b c \cos A=0 \\
\Rightarrow 25 c \cos A=9 b \Rightarrow 25 \cos A \sin C=9 \sin B=9 \sin (A+C) \\
\Rightarrow 16 \cos A \sin C=9 \sin A \cos C \Rightarrow \tan A=\frac{16}{9} \tan C,
\end{array}
$$

Thus, $\tan (A-C)=\frac{\tan A-\tan C}{1+\tan A \cdot \tan C}=\frac{\frac{7}{9} \tan C}{1+\frac{16}{9} \tan ^{2} C}=\frac{7 \tan C}{9+16 \tan ^{2} C}$
$=\frac{7}{\frac{9}{\tan C}+16 \tan C} \leqslant \frac{7}{2 \sqrt{9 \cdot 16}}=\frac{7}{24}$, equality holds when $\tan C=\frac{3}{4} \Rightarrow \tan A=\frac{4}{3}$.
Since $\sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}=\frac{24}{25}, \cos 2 A=\frac{1-\tan ^{2} A}{1+\tan ^{2} A}=-\frac{7}{25}$,
therefore $\sin (2 A-C)=\sin 2 A \cos C-\cos 2 A \sin C=\frac{24}{25} \cdot \frac{4}{5}+\frac{7}{25} \cdot \frac{3}{5}=\frac{117}{125}$.