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

A

$$\frac{1 - \sqrt{3}}{\sqrt{2}}$$

B

$$\frac{3 - 2\sqrt{2}}{2\sqrt{3}}$$

C

$$\frac{3 - 2\sqrt{2}}{3\sqrt{2}}$$

D

$$\frac{1 - 2\sqrt{2}}{\sqrt{3}}$$