30°

Since $\angle EDA=60°$ , $\angle EDC=90°-60°=30°$ , and it is clear that \angle $DCE=90°+30°=120°.$ From that $\angle CED=180°-120°-30°=30°$ .

So $\triangle CDE$ is isosceles, since it has two equal angles. $ABCD$ is a rectangle, so $AB=12=CD=CE$ . Note that $\triangle BCE$ is a half-equilateral triangle, because if we reflect $E$ to $B$ , the $\triangle E'CE$ will be equilateral. So $CE=2*BE=12$ .

Therefre $\overline{BE}=\boxed{6}$ .