7. Given the circles $O_{1}: x^{2}+y^{2}=2$ and $O_{2}:(x-3)^{2}+y^{2}=5$ intersect at point $A$ in the first quadrant, a line through point $A$ intersects circles $O_{1}, O_{2}$ at points $C, D$ respectively $(C, D$ distinct from point $A)$, and $\overrightarrow{A C}=2 \overrightarrow{A D}$, then the slope of line $C D$ is
As shown in the figure, $E, F$ are the midpoints of $A C, A D$ respectively, and it is easy to know that $A(1,1)$,
Let $C D: y=k(x-1)+1$,
$\left|O_{1} E\right|=\frac{|-k+1|}{\sqrt{k^{2}+1}},\left|O_{2} F\right|=\frac{|2 k+1|}{\sqrt{k^{2}+1}}$,
Then $|A E|^{2}=2-\frac{k^{2}-2 k+1}{k^{2}+1}=\frac{k^{2}+2 k+1}{k^{2}+1}$,
$$
|A F|^{2}=5-\frac{4 k^{2}+4 k+1}{k^{2}+1}=\frac{k^{2}-4 k+4}{k^{2}+1} \text {. }
$$

According to the problem, $|A E|=2|A F| \Rightarrow k^{2}+2 k+1=4\left(k^{2}-4 k+4\right) \Rightarrow k^{2}-6 k+5=0$ $\Rightarrow k=5$ or $k=1$. Combining with the graph, $k=1$ should be discarded. Therefore, the slope of the line $C D$ is 5.