7. Given $I$ is the incenter of $\triangle A B C$, and
$$
9 \overrightarrow{C I}=4 \overrightarrow{C A}+3 \overrightarrow{C B} \text {. }
$$

Let $R$ and $r$ be the circumradius and inradius of $\triangle A B C$, respectively. Then $\frac{r}{R}=$ $\qquad$ .
7. $\frac{5}{16}$.

The equation simplifies to
$$
4 \overrightarrow{I A}+3 \overrightarrow{I B}+2 \overrightarrow{I C}=\mathbf{0} \text {. }
$$

Assume the side lengths of $\triangle A B C$ are $a=4, b=3$, $c=2$.
Then $\sin A=\frac{\sqrt{15}}{4}$.
Using the Law of Sines, we get $R=\frac{8}{\sqrt{15}}$.
By $\frac{1}{2}(a+b+c) r=\frac{1}{2} b c \sin A$
$$
\Rightarrow r=\frac{\sqrt{15}}{6} \text {. }
$$

Therefore, $\frac{r}{R}=\frac{5}{16}$.