Let $$P(x_1, y_1)$$ and $$Q(x_2, y_2)$$ be two distinct points on the ellipse

$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$

such that $$y_1 > 0$$, and $$y_2 > 0$$. Let $$\mathcal{C}$$ denote the circle $$x^2 + y^2 = 9$$, and $$M$$ be the point $$(3, 0)$$.

Suppose the line $$x = x_1$$ intersects $$\mathcal{C}$$ at $$R$$, and the line $$x = x_2$$ intersects $$\mathcal{C}$$ at $$S$$, such that the $$y$$-coordinates of $$R$$ and $$S$$ are positive. Let $$\angle ROM = \frac{\pi}{6}$$ and $$\angle SOM = \frac{\pi}{3}$$, where $$O$$ denotes the origin $$(0, 0)$$. Let $$|XY|$$ denote the length of the line segment $$XY$$.

Then which of the following statements is (are) TRUE?

The ellipse is $$\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$$ and the circle $$\mathcal{C}$$ is $$x^{2}+y^{2}=9$$ (radius $$3$$, centre at the origin $$O(0,0)$$).
The point $$M$$ is $$(3,0)$$, so the ray $$OM$$ is the positive $$x$$-axis.

Because $$\angle ROM=\frac{\pi}{6}$$ and $$|OR|=3$$, the point $$R$$ on the circle is obtained by standard polar-Cartesian conversion:

$$R=\bigl(3\cos\frac{\pi}{6},\,3\sin\frac{\pi}{6}\bigr)=\left(\frac{3\sqrt{3}}{2},\,\frac{3}{2}\right).$$

Similarly, $$\angle SOM=\frac{\pi}{3}$$ gives

$$S=\bigl(3\cos\frac{\pi}{3},\,3\sin\frac{\pi}{3}\bigr)=\left(\frac{3}{2},\,\frac{3\sqrt{3}}{2}\right).$$

The vertical lines through $$R$$ and $$S$$ intersect the ellipse at $$P$$ and $$Q$$, so

$$x_{1}=x_{R}=\frac{3\sqrt{3}}{2},\qquad x_{2}=x_{S}=\frac{3}{2}.$$

Point P
Insert $$x_{1}$$ in the ellipse:

$$\frac{x_{1}^{2}}{9}=\frac{\left(\dfrac{3\sqrt{3}}{2}\right)^{2}}{9}=\frac{27/4}{9}=\frac34.$$

Therefore $$\frac{y_{1}^{2}}{4}=1-\frac34=\frac14 \;\Longrightarrow\; y_{1}=1 \;(\text{positive branch}).$$

Hence $$P=\left(\frac{3\sqrt{3}}{2},\,1\right).$$

Point Q
Insert $$x_{2}$$ in the ellipse:

$$\frac{x_{2}^{2}}{9}=\frac{\left(\dfrac{3}{2}\right)^{2}}{9}=\frac{9/4}{9}=\frac14.$$

Therefore $$\frac{y_{2}^{2}}{4}=1-\frac14=\frac34 \;\Longrightarrow\; y_{2}=\sqrt{3}.$$

Hence $$Q=\left(\frac{3}{2},\,\sqrt{3}\right).$$

Equation of line $$PQ$$
Slope $$m=\dfrac{\sqrt{3}-1}{\frac{3}{2}-\frac{3\sqrt{3}}{2}} =\dfrac{\sqrt{3}-1}{\frac{3}{2}(1-\sqrt{3})} =-\frac{2}{3}.$$

Using point $$P$$:

$$y-1=-\frac{2}{3}\Bigl(x-\frac{3\sqrt{3}}{2}\Bigr) \;\Longrightarrow\;3y-3=-2x+3\sqrt{3}.$$

Rearranging,

$$2x+3y=3(1+\sqrt{3}).$$

This matches Option A. Option B (with $$2x+y$$) is therefore incorrect.

Checking Statement C
Take $$N_{2}=(x_{2},0)=\left(\dfrac32,0\right).$$
Because $$Q$$ and $$S$$ have the same $$x$$-coordinate $$x_{2},$$ the required distances are vertical:

$$|N_{2}Q|=|y_{2}|=\sqrt{3},\qquad |N_{2}S|=\left|\frac{3\sqrt{3}}{2}\right|=\frac{3\sqrt{3}}{2}.$$

Thus $$3|N_{2}Q|=3\sqrt{3},\qquad 2|N_{2}S|=2\cdot\frac{3\sqrt{3}}{2}=3\sqrt{3},$$ so $$3|N_{2}Q|=2|N_{2}S|.$$
Statement C is true.

Checking Statement D
Take $$N_{1}=(x_{1},0)=\left(\dfrac{3\sqrt{3}}{2},0\right).$$

Vertical distances give $$|N_{1}P|=1,\quad |N_{1}R|=\frac{3}{2}.$$

But $$9|N_{1}P|=9,\quad 4|N_{1}R|=4\cdot\frac32=6,$$ which are not equal. Hence Statement D is false.

Final result
Option A and Option C are the only correct statements.

Correct choices: Option A, Option C.