A rotation by $120^{\circ}$ about the center of the circle will take $ABE$ to $BCD$, so $BCD$ has area 3. Let $AD=x, BD=y$, and observe that $\angle ADC=\angle CDB=60^{\circ}$. By Ptolemy's Theorem, $CD=x+y$. We have $$4=[ACD]=\frac{1}{2} AD \cdot CD \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{4} x(x+y)$$ $$3=[BCD]=\frac{1}{2} BD \cdot CD \cdot \sin 60^{\circ}=\frac{\sqrt{3}}{4} y(x+y)$$ By dividing these equations find $x: y=4: 3$. Let $x=4t, y=3t$. Substitute this into the first equation to get $1=\frac{\sqrt{3}}{4} \cdot 7t^{2}$. By the Law of Cosines, $$AB^{2}=x^{2}+xy+y^{2}=37t^{2}$$ The area of $ABC$ is then $$\frac{AB^{2} \sqrt{3}}{4}=\frac{37}{7}$$