Let the ellipse $\mathrm{E}_1: \frac{x^2}{\mathrm{a}^2} + \frac{y^2}{\mathrm{b}^2} = 1,\ \mathrm{a} > \mathrm{b}$ and $\mathrm{E}_2: \frac{x^2}{\mathrm{A}^2} + \frac{y^2}{\mathrm{B}^2} = 1,\ \mathrm{A} < \mathrm{B}$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be 4. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $ABCD$ equals: (1) $\frac{12\sqrt{6}}{5}$ (2) $6\sqrt{6}$ (3) $\frac{18\sqrt{6}}{5}$ (4) $\frac{24\sqrt{6}}{5}$
Let the ellipse $\mathrm{E}_1: \frac{x^2}{\mathrm{a}^2} + \frac{y^2}{\mathrm{b}^2} = 1,\ \mathrm{a} > \mathrm{b}$ and $\mathrm{E}_2: \frac{x^2}{\mathrm{A}^2} + \frac{y^2}{\mathrm{B}^2} = 1,\ \mathrm{A} < \mathrm{B}$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be 4. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $ABCD$ equals:\\
(1) $\frac{12\sqrt{6}}{5}$\\
(2) $6\sqrt{6}$\\
(3) $\frac{18\sqrt{6}}{5}$\\
(4) $\frac{24\sqrt{6}}{5}$