Let the tangent and normal at the point $$(3\sqrt{3}, 1)$$ on the ellipse $$\frac{x^2}{36} + \frac{y^2}{4} = 1$$ meet the $$y-$$axis at the points $$A$$ and $$B$$ respectively. Let the circle $$C$$ be drawn taking $$AB$$ as a diameter and the line $$x = 2\sqrt{5}$$ intersect $$C$$ at the points P and Q. If the tangents at the points P and Q on the circle intersect at the point $$(\alpha, \beta)$$, then $$\alpha^2 - \beta^2$$ is equal to

Given: ellipse $$\frac{x^2}{36} + \frac{y^2}{4} = 1$$, point $$(3\sqrt{3}, 1)$$.

Verify point on ellipse: $$\frac{27}{36} + \frac{1}{4} = \frac{3}{4} + \frac{1}{4} = 1$$. ✓

Tangent at $$(3\sqrt{3}, 1)$$:

$$\frac{3\sqrt{3} \cdot x}{36} + \frac{y}{4} = 1 \implies \frac{\sqrt{3}x}{12} + \frac{y}{4} = 1 \implies \sqrt{3}x + 3y = 12$$

At $$y$$-axis ($$x = 0$$): $$y = 4$$. So $$A = (0, 4)$$.

Normal at $$(3\sqrt{3}, 1)$$:

Tangent slope: $$y' = -\frac{4x}{36y} = -\frac{4 \cdot 3\sqrt{3}}{36} = -\frac{\sqrt{3}}{3}$$. Normal slope $$= \sqrt{3}$$.

$$y - 1 = \sqrt{3}(x - 3\sqrt{3}) \implies y = \sqrt{3}x - 8$$

At $$y$$-axis: $$y = -8$$. So $$B = (0, -8)$$.

Circle $$C$$ with diameter $$AB$$:

Center $$= (0, -2)$$, radius $$= 6$$. Equation: $$x^2 + (y + 2)^2 = 36$$.

Intersection with $$x = 2\sqrt{5}$$:

$$20 + (y + 2)^2 = 36 \implies (y + 2)^2 = 16 \implies y = 2$$ or $$y = -6$$.

$$P = (2\sqrt{5}, 2)$$ and $$Q = (2\sqrt{5}, -6)$$.

Tangents at $$P$$ and $$Q$$:

Tangent at $$P(2\sqrt{5}, 2)$$: $$2\sqrt{5} \cdot x + (2 + 2)(y + 2) = 36 \implies 2\sqrt{5}x + 4y = 28$$

Tangent at $$Q(2\sqrt{5}, -6)$$: $$2\sqrt{5} \cdot x + (-6 + 2)(y + 2) = 36 \implies 2\sqrt{5}x - 4y = 44$$

Intersection point $$(\alpha, \beta)$$:

Adding: $$4\sqrt{5}x = 72 \implies x = \frac{18}{\sqrt{5}} = \frac{18\sqrt{5}}{5}$$

Subtracting: $$8y = -16 \implies y = -2$$

So $$\alpha = \frac{18\sqrt{5}}{5}$$, $$\beta = -2$$.

$$\alpha^2 - \beta^2 = \frac{324 \times 5}{25} - 4 = \frac{1620}{25} - 4 = \frac{324}{5} - \frac{20}{5} = \boxed{\frac{304}{5}}$$

The correct answer is Option C.