3. In $\triangle A B C$, $A B=2, A C=1, B C=\sqrt{7}$, $O$ is the circumcenter of $\triangle A B C$, and $\overrightarrow{A O}=\lambda \overrightarrow{A B}+\mu \overrightarrow{A C}$. Then $\lambda+\mu=$ $\qquad$ .
3. $\frac{13}{6}$.

In $\triangle A B C$, by the cosine rule we have
$$
\angle B A C=120^{\circ} \Rightarrow \overrightarrow{A B} \cdot \overrightarrow{A C}=-1 \text {. }
$$

Also, $O$ is the circumcenter of $\triangle A B C$, so
$$
\begin{array}{l}
\left\{\begin{array}{l}
\overrightarrow{A O} \cdot \overrightarrow{A B}=\frac{1}{2} \overrightarrow{A B}^{2}=2, \\
\overrightarrow{A O} \cdot \overrightarrow{A C}=\frac{1}{2} \overrightarrow{A C}^{2}=\frac{1}{2}
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
2=\overrightarrow{A O} \cdot \overrightarrow{A B}=\lambda \overrightarrow{A B}^{2}+\mu \overrightarrow{A C} \cdot \overrightarrow{A B}=4 \lambda-\mu, \\
\frac{1}{2}=\overrightarrow{A O} \cdot \overrightarrow{A C}=\lambda \overrightarrow{A B} \cdot \overrightarrow{A C}+\mu \overrightarrow{A C^{2}}=-\lambda+\mu
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\lambda=\frac{5}{6}, \\
\mu=\frac{8}{6}
\end{array} \Rightarrow \lambda+\mu=\frac{13}{6} .\right.
\end{array}
$$