11. As shown in Figure 4, in $\triangle A B C$, $\angle B A C=150^{\circ}$, $B C=74$, point $D$ is on side $B C$ such that $B D=14$. If $\angle A D B=60^{\circ}$, then the area of $\triangle A B C$ is $\qquad$.
11. $111 \sqrt{3}$.

As shown in Figure 9, let $B F$ and $C F$ be the lines symmetric to $B C$ with respect to $A B$ and $C A$, respectively, intersecting at point $F$. Point $D$ is symmetric to point $E$ on $C F$, and $A F$ is connected.
Thus, $A$ is the incenter of $\triangle F B C$.
Since $\angle B A C=150^{\circ}$, we have
$$
\begin{array}{l}
\angle A B C+\angle A C B=30^{\circ}, \\
\angle F B C+\angle F C B=60^{\circ}, \\
\angle B F C=120^{\circ}, \angle A F E=60^{\circ} .
\end{array}
$$

Also, $\angle A E F=\angle A D B=60^{\circ}$, so $\triangle A E F$ is an equilateral triangle.
By the Angle-Angle-Side (AAS) theorem, we know that $\triangle A B F \cong \triangle A B D$.
Let $E F=A F=A E=A D=x$.
Given $B F=B D=14, C E=C D=60$, applying the Law of Cosines in $\triangle F B C$ yields
$$
\begin{array}{l}
14^{2}+(60+x)^{2}-74^{2}=2 \times 14(60+x) \frac{1}{2} \\
\Rightarrow x=6 .
\end{array}
$$

Therefore, $S_{\triangle A B C}=\frac{1}{2} \times 74 \times 6 \sin 60^{\circ}=111 \sqrt{3}$.