2. In a convex pentagon $A B C D E$, it is known that $D E / / A C, \angle A D C=90^{\circ}, A C=12, C D=6, A C$ bisects $\angle D A B, \angle B C E=60^{\circ}$. Then the minimum value of the area of $\triangle B C E$ is $\qquad$
Given, $\angle C D E=120^{\circ}, \angle C A B=30^{\circ}$, and $\angle D C A=$
$\angle E C B=60^{\circ} \Rightarrow \angle D C E=\angle A C B$, then in $\triangle D C E$
and $\triangle A C B$, $\frac{x}{\sin 120^{\circ}}=\frac{6}{\sin \left(60^{\circ}-\theta\right)}$,
$\frac{y}{\sin 30^{\circ}}=\frac{12}{\sin \left(150^{\circ}-\theta\right)}$, let $\alpha=60^{\circ}-\theta$,
thus $\sin \alpha=\frac{3 \sqrt{3}}{x}, \cos \alpha=\frac{6}{y} \Rightarrow \frac{18 \sqrt{3}}{x y} \leqslant \frac{1}{2} \Rightarrow x y \geqslant 36 \sqrt{3}$, equality holds when $\theta=15^{\circ}$.
Therefore, the minimum area of $\triangle B C E$ is $\frac{1}{2} \cdot 36 \sqrt{3} \cdot \frac{\sqrt{3}}{2}=27$.