7. Let $O$ be the circumcenter of $\triangle A B C$. If $\overrightarrow{A O}=\overrightarrow{A B}+2 \overrightarrow{A C}$, then the value of $\sin \angle B A C$ is
As shown in the figure, take $\overrightarrow{A D}=\frac{1}{3} \overrightarrow{A O}=\frac{1}{3} \overrightarrow{A B}+\frac{2}{3} \overrightarrow{A C} \Rightarrow B, D, C$ are collinear and $\overrightarrow{B D}=\frac{2}{3} \overrightarrow{B C}$. Let $C D=x, A D=y \Rightarrow B D=2 x, O D=2 y$, then the circumradius $R=3 y$ of $\triangle A B C$. In $\triangle A B C$, by the Law of Sines, we have $\frac{B C}{\sin \angle B A C}=2 R$
$$
\Rightarrow \frac{3 x}{\sin \angle B A C}=6 y \Rightarrow \sin \angle B A C=\frac{x}{2 y} \text {. }
$$

Then, applying the Law of Cosines in $\triangle O C D$ and $\triangle O B D$, we get
$$
\begin{array}{l}
\frac{x^{2}+4 y^{2}-9 y^{2}}{4 x y}=-\frac{4 x^{2}+4 y^{2}-9 y^{2}}{8 x y} \\
\Rightarrow 2\left(x^{2}-5 y^{2}\right)=-\left(4 x^{2}-5 y^{2}\right) \Rightarrow 2 x^{2}=5 y^{2} \\
\Rightarrow \sin ^{2} \angle B A C=\frac{x^{2}}{4 y^{2}}=\frac{5}{8} \Rightarrow \sin \angle B A C=\frac{\sqrt{10}}{4} .
\end{array}
$$