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

Given: $$\frac{dy}{dx} + \frac{x+a}{y-2} = 0$$, $$y(1) = 0$$, and the enclosed area is $$4\pi$$.

Rearranging: $$(y-2)\,dy = -(x+a)\,dx$$

Integrating both sides:

$$ \frac{(y-2)^2}{2} + \frac{(x+a)^2}{2} = C $$

This represents a circle: $$(x+a)^2 + (y-2)^2 = 2C$$ with center $$(-a, 2)$$ and radius $$r = \sqrt{2C}$$.

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

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

The area of the circle is $$\pi r^2 = 2\pi C = 4\pi$$, so $$C = 2$$ and $$r = 2$$.

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

The circle is $$(x-1)^2 + (y-2)^2 = 4$$, centered at $$(1, 2)$$ with radius 2.

Intersection with y-axis ($$x = 0$$):

$$ 1 + (y-2)^2 = 4 \Rightarrow (y-2)^2 = 3 \Rightarrow y = 2 \pm \sqrt{3} $$

$$P(0, 2+\sqrt{3})$$ and $$Q(0, 2-\sqrt{3})$$.

Normal at P: $$\frac{dy}{dx} = -\frac{x-1}{y-2}$$. At P: slope = $$\frac{1}{\sqrt{3}}$$. Normal slope = $$-\sqrt{3}$$.

Normal: $$y - (2+\sqrt{3}) = -\sqrt{3}(x-0)$$. At $$y = 0$$: $$x = \frac{2+\sqrt{3}}{\sqrt{3}} = \frac{2\sqrt{3}}{3} + 1$$

Normal at Q: Slope at Q = $$\frac{1}{-\sqrt{3}}$$. Normal slope = $$\sqrt{3}$$.

Normal: $$y - (2-\sqrt{3}) = \sqrt{3}x$$. At $$y = 0$$: $$x = \frac{-(2-\sqrt{3})}{\sqrt{3}} = \frac{\sqrt{3}-2}{\sqrt{3}} = 1 - \frac{2\sqrt{3}}{3}$$

Length RS:

$$ RS = \left|\frac{2\sqrt{3}}{3} + 1 - 1 + \frac{2\sqrt{3}}{3}\right| = \frac{4\sqrt{3}}{3} $$

Therefore, $$RS = \frac{4\sqrt{3}}{3}$$.