Let the product of the focal distances of the point $\left(\sqrt{3}, \frac{1}{2}\right)$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $(a > b)$, be $\frac{7}{4}$. Then the absolute difference of the eccentricities of two such ellipses is\\
(1) $\frac{1 - \sqrt{3}}{\sqrt{2}}$\\
(2) $\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$\\
(3) $\frac{3 - 2\sqrt{2}}{3\sqrt{2}}$\\
(4) $\frac{1 - 2\sqrt{2}}{\sqrt{3}}$