9. (16 points) Let the function $f(x)=3 a x^{2}-2 b x+c$. If $a-b+c=0, f(0)>0, f(1)>0$, find the range of $\frac{a+3 b+7 c}{2 a+b}$.

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$$
\begin{array}{l}
f(0)=c>0, \\
f(1)=3 a-2 b+c>0, \\
a-b+c=0 .
\end{array}
$$

From equations (1), (3) and equations (2), (3), we get
$$
ab \Rightarrow 2 a>b>a \text {. }
$$

Substituting $b=a+c$ into equation (2) yields
$$
a>c \Rightarrow a>0 \text {. }
$$

Then, from $2 a>b$, we get $1<\frac{b}{a}<2$.
$$
\begin{array}{l}
\text { Therefore, } \frac{a+3 b+7 c}{2 a+b}=\frac{10 b-6 a}{2 a+b} \\
=10-\frac{26}{\frac{b}{a}+2} \in\left(\frac{4}{3}, \frac{7}{2}\right) .
\end{array}
$$