The area enclosed by the closed curve $$C$$ given by the differential equation $$\frac{dy}{dx} + \frac{x+a}{y-2} = 0$$, $$y(1) = 0$$ is $$4\pi$$. Let $$P$$ and $$Q$$ be the points of intersection of the curve $$C$$ and the $$y$$-axis. If normals at $$P$$ and $$Q$$ on the curve $$C$$ intersect $$x$$-axis at points $$R$$ and $$S$$ respectively, then the length of the line segment $$RS$$ is

$$\frac{dy}{dx} + \frac{x+a}{y-2} = 0 \implies (y-2)dy = -(x+a)dx$$

$$\int (y-2)dy = -\int (x+a)dx \implies \frac{(y-2)^2}{2} = -\frac{(x+a)^2}{2} + k_1$$

$$(x+a)^2 + (y-2)^2 = r^2$$

Using given area enclosed by the closed curve $$C = 4\pi$$:

$$\pi r^2 = 4\pi \implies r^2 = 4 \implies r = 2$$

Using initial condition $$y(1) = 0$$:

$$(1+a)^2 + (0-2)^2 = 4 \implies (1+a)^2 + 4 = 4 \implies 1+a = 0 \implies a = -1$$

$$\text{Curve } C: (x-1)^2 + (y-2)^2 = 4 \implies \text{Center } O(1, 2)$$

Finding points of intersection $$P$$ and $$Q$$ with $$y$$-axis ($$x=0$$):

$$(0-1)^2 + (y-2)^2 = 4 \implies 1 + (y-2)^2 = 4 \implies (y-2)^2 = 3$$

$$y - 2 = \pm\sqrt{3} \implies y = 2 \pm \sqrt{3}$$

$$P = (0, \ 2 + \sqrt{3}), \quad Q = (0, \ 2 - \sqrt{3})$$

Since the normal at any point on a circle passes through its center:

$$\text{Normals at } P \text{ and } Q \text{ intersect at center } O(1, 2)$$

Determining lines $$R$$ and $$S$$ separately using standard normal slopes:

$$m_{OP} = \frac{2+\sqrt{3}-2}{0-1} = -\sqrt{3} \implies \text{Equation of normal at } P: y - 2 = -\sqrt{3}(x - 1)$$

$$m_{OQ} = \frac{2-\sqrt{3}-2}{0-1} = \sqrt{3} \implies \text{Equation of normal at } Q: y - 2 = \sqrt{3}(x - 1)$$

Finding $$x$$-intercepts $$R$$ and $$S$$ ($$y=0$$):

$$0 - 2 = -\sqrt{3}(x_R - 1) \implies x_R - 1 = \frac{2}{\sqrt{3}} \implies x_R = 1 + \frac{2}{\sqrt{3}}$$

$$0 - 2 = \sqrt{3}(x_S - 1) \implies x_S - 1 = -\frac{2}{\sqrt{3}} \implies x_S = 1 - \frac{2}{\sqrt{3}}$$

$$R = \left(1 + \frac{2}{\sqrt{3}}, \ 0\right), \quad S = \left(1 - \frac{2}{\sqrt{3}}, \ 0\right)$$

Calculating length of line segment $$RS$$:

$$RS = \vert{}x_R - x_S\vert{} = \left\vert{}\left(1 + \frac{2}{\sqrt{3}}\right) - \left(1 - \frac{2}{\sqrt{3}}\right)\right\vert{} = \frac{4}{\sqrt{3}} = \frac{4\sqrt{3}}{3}$$