11. Given $O$ is the circumcenter of $\triangle A B C$, $|A B|=2$, $|A C|=1, \angle B A C=\frac{2 \pi}{3}$. Let $\boldsymbol{A B}=\boldsymbol{a}, \boldsymbol{A} \boldsymbol{C}=\boldsymbol{b}$. If $\boldsymbol{A} \boldsymbol{O}=\lambda_{1} \boldsymbol{a}+\lambda_{2} \boldsymbol{b}$, then $\lambda_{1}+\lambda_{2}=$
11. $\frac{13}{6}$.

Establish a rectangular coordinate system as shown in Figure 3. Then $A(0,0)$, $B(2, 0)$, and $C\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$.
Obviously, the midpoint of $A C$ is $M\left(-\frac{1}{4}, \frac{\sqrt{3}}{4}\right)$. Let $O(1, y)$.
Since $O M \perp A C$, we have $O M \cdot A C=0$, which means $-\frac{5}{4} \times\left(-\frac{1}{2}\right)+\left(\frac{\sqrt{3}}{4}-y\right) \times \frac{\sqrt{3}}{2}=0$.
Solving this, we get $y=\frac{2 \sqrt{3}}{3}$, and $A O=\left(1, \frac{2 \sqrt{3}}{3}\right)$.
Given $A O=\lambda_{1} a+\lambda_{2} b$, we have
$1=2 \lambda_{1}+\left(-\frac{1}{2}\right) \lambda_{2}$

and $\frac{2 \sqrt{3}}{3}=\frac{\sqrt{3}}{2} \lambda_{2}$.
Solving these, we get $\lambda_{2}=\frac{4}{3}$, $\lambda_{1}=\frac{5}{6}$.
Thus, $\lambda_{1}+\lambda_{2}=\frac{13}{6}$.