Let an ellipse $$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, $$a^2 > b^2$$, passes through $$\left(\sqrt{\frac{3}{2}}, 1\right)$$ and has eccentricity $$\frac{1}{\sqrt{3}}$$. If a circle, centered at focus $$F(\alpha, 0)$$, $$\alpha > 0$$, of $$E$$ and radius $$\frac{2}{\sqrt{3}}$$, intersects $$E$$ at two points $$P$$ and $$Q$$, then $$PQ^2$$ is equal to:

We are given the ellipse $$E: \dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$$ with $$a^2 > b^2$$, passing through $$\left(\sqrt{\dfrac{3}{2}},\, 1\right)$$ and having eccentricity $$e = \dfrac{1}{\sqrt{3}}$$.

Step 1: Find $$a^2$$ and $$b^2$$.

From the eccentricity relation:

$$e^2 = 1 - \frac{b^2}{a^2} \implies \frac{1}{3} = 1 - \frac{b^2}{a^2} \implies b^2 = \frac{2a^2}{3}$$

Since $$\left(\sqrt{\dfrac{3}{2}},\, 1\right)$$ lies on the ellipse:

$$\frac{3/2}{a^2} + \frac{1}{b^2} = 1$$

Substituting $$b^2 = \dfrac{2a^2}{3}$$:

$$\frac{3}{2a^2} + \frac{3}{2a^2} = 1 \implies \frac{3}{a^2} = 1 \implies a^2 = 3$$

Hence $$b^2 = 2$$, and the ellipse is $$\dfrac{x^2}{3} + \dfrac{y^2}{2} = 1$$.

Step 2: Identify the focus and circle.

$$c = \sqrt{a^2 - b^2} = \sqrt{3 - 2} = 1$$

The focus with positive $$x$$-coordinate is $$F(1, 0)$$. The circle centered at $$F$$ with radius $$\dfrac{2}{\sqrt{3}}$$ has equation:

$$(x - 1)^2 + y^2 = \frac{4}{3}$$

Step 3: Find the intersection points $$P$$ and $$Q$$.

From the ellipse: $$y^2 = 2 - \dfrac{2x^2}{3}$$

Substituting into the circle equation:

$$(x - 1)^2 + 2 - \frac{2x^2}{3} = \frac{4}{3}$$

$$x^2 - 2x + 1 + 2 - \frac{2x^2}{3} = \frac{4}{3}$$

$$\frac{x^2}{3} - 2x + 3 = \frac{4}{3}$$

$$x^2 - 6x + 9 = 4$$

$$x^2 - 6x + 5 = 0$$

$$(x - 1)(x - 5) = 0$$

So $$x = 1$$ or $$x = 5$$. For $$x = 5$$: $$y^2 = 2 - \dfrac{50}{3} < 0$$, which is invalid. Thus $$x = 1$$.

At $$x = 1$$: $$y^2 = 2 - \dfrac{2}{3} = \dfrac{4}{3}$$, giving $$y = \pm\dfrac{2}{\sqrt{3}}$$.

So $$P = \left(1,\, \dfrac{2}{\sqrt{3}}\right)$$ and $$Q = \left(1,\, -\dfrac{2}{\sqrt{3}}\right)$$.

Step 4: Calculate $$PQ^2$$.

$$PQ = \frac{2}{\sqrt{3}} + \frac{2}{\sqrt{3}} = \frac{4}{\sqrt{3}}$$

$$PQ^2 = \frac{16}{3}$$

The correct answer is Option C: $$\dfrac{16}{3}$$.