Let P be a point on the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$. Let the line passing through P and parallel to y-axis meet the circle $$x^2 + y^2 = 9$$ at point Q such that P and Q are on the same side of the x-axis. Then, the eccentricity of the locus of the point R on PQ such that $$PR : RQ = 4 : 3$$ as P moves on the ellipse, is:

Let P be a point on the ellipse $$\frac{x^2}{9} + \frac{y^2}{4} = 1$$. Parametrize P using the parametric equations for the ellipse: $$x = 3\cos\theta$$, $$y = 2\sin\theta$$, so P is $$(3\cos\theta, 2\sin\theta)$$.

The line passing through P and parallel to the y-axis has the equation $$x = 3\cos\theta$$. This line intersects the circle $$x^2 + y^2 = 9$$. Substitute $$x = 3\cos\theta$$ into the circle equation:

$$(3\cos\theta)^2 + y^2 = 9$$

$$9\cos^2\theta + y^2 = 9$$

$$y^2 = 9 - 9\cos^2\theta = 9\sin^2\theta$$

$$y = \pm 3|\sin\theta|$$

Since P and Q are on the same side of the x-axis, the y-coordinate of Q must have the same sign as that of P. For P, the y-coordinate is $$2\sin\theta$$, so when $$\sin\theta \geq 0$$, Q has a non-negative y-coordinate, and when $$\sin\theta < 0$$, Q has a negative y-coordinate. In both cases, Q can be taken as $$(3\cos\theta, 3\sin\theta)$$.

Thus, Q is $$(3\cos\theta, 3\sin\theta)$$.

Point R is on PQ such that $$PR : RQ = 4 : 3$$. Since PQ is a vertical line segment (same x-coordinate), R divides PQ in the ratio $$PR : RQ = 4 : 3$$. Using the section formula, if R divides the line segment joining $$P(x_1, y_1)$$ and $$Q(x_2, y_2)$$ in the ratio $$m : n$$, then:

$$R = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right)$$

Here, $$m = 4$$, $$n = 3$$, $$P(3\cos\theta, 2\sin\theta)$$, $$Q(3\cos\theta, 3\sin\theta)$$.

The x-coordinate of R is:

$$x_R = \frac{4 \cdot (3\cos\theta) + 3 \cdot (3\cos\theta)}{7} = \frac{12\cos\theta + 9\cos\theta}{7} = \frac{21\cos\theta}{7} = 3\cos\theta$$

The y-coordinate of R is:

$$y_R = \frac{4 \cdot (3\sin\theta) + 3 \cdot (2\sin\theta)}{7} = \frac{12\sin\theta + 6\sin\theta}{7} = \frac{18\sin\theta}{7}$$

So R is $$(3\cos\theta, \frac{18}{7} \sin\theta)$$.

To find the locus of R as $$\theta$$ varies, eliminate $$\theta$$:

Set $$x = 3\cos\theta$$ and $$y = \frac{18}{7} \sin\theta$$. Then:

$$\frac{x}{3} = \cos\theta, \quad \frac{7y}{18} = \sin\theta$$

Squaring and adding:

$$\left( \frac{x}{3} \right)^2 + \left( \frac{7y}{18} \right)^2 = \cos^2\theta + \sin^2\theta = 1$$

$$\frac{x^2}{9} + \frac{49y^2}{324} = 1$$

This is the equation of an ellipse. Rewriting it in standard form $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$:

$$\frac{x^2}{9} + \frac{y^2}{\frac{324}{49}} = 1$$

So $$a^2 = 9$$ and $$b^2 = \frac{324}{49}$$. Since $$a^2 > b^2$$ (as $$9 > \frac{324}{49} \approx 6.612$$), the major axis is along the x-axis.

The eccentricity $$e$$ is given by:

$$e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{\frac{324}{49}}{9}} = \sqrt{1 - \frac{324}{49 \times 9}} = \sqrt{1 - \frac{324}{441}}$$

Simplify $$\frac{324}{441}$$:

$$\frac{324 \div 9}{441 \div 9} = \frac{36}{49}$$

So:

$$e = \sqrt{1 - \frac{36}{49}} = \sqrt{\frac{49 - 36}{49}} = \sqrt{\frac{13}{49}} = \frac{\sqrt{13}}{7}$$

The eccentricity is $$\frac{\sqrt{13}}{7}$$, which corresponds to option D.