7. Given that $O$ is the circumcenter of $\triangle A B C$. If $A B=A C$, $\angle C A B=30^{\circ}$, and $\overrightarrow{C O}=\lambda_{1} \overrightarrow{C A}+\lambda_{2} \overrightarrow{C B}$, then $\lambda_{1} \lambda_{2}=$ $\qquad$ .
7. $7 \sqrt{3}-12$.

Let's assume $AB=2$. Establish a Cartesian coordinate system with $A$ as the origin and the line $AB$ as the $x$-axis. Then
$$
A(0,0), B(2,0), C(\sqrt{3}, 1) \text {. }
$$

Let the circumcenter be $O(1, y)$.
From $|\overrightarrow{O A}|=|\overrightarrow{O C}|$, we get
$$
1+y^{2}=(\sqrt{3}-1)^{2}+(y-1)^{2} \text {. }
$$

Solving this, we get $y=2-\sqrt{3}$.
$$
\begin{array}{l}
\text { Then }(1-\sqrt{3}, 1-\sqrt{3})=\overrightarrow{C O}=\lambda_{1} \overrightarrow{C A}+\lambda_{2} \overrightarrow{C B} \\
=\lambda_{1}(-\sqrt{3},-1)+\lambda_{2}(2-\sqrt{3},-1) .
\end{array}
$$

Solving this, we get $\lambda_{1}=2 \sqrt{3}-3, \lambda_{2}=2-\sqrt{3}$.
$$
\text { Therefore } \lambda_{1} \lambda_{2}=7 \sqrt{3}-12 \text {. }
$$