Let $$H : \frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4\sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha > 0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2 + \beta$$ is equal to

Hyperbola $$H: -\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$, eccentricity $$e = \sqrt{3}$$, latus rectum $$\frac{2a^2}{b} = 4\sqrt{3}$$.

$$e^2 = 1 + \frac{a^2}{b^2} \implies 3 = 1 + \frac{a^2}{b^2} \implies a^2 = 2b^2$$.

From latus rectum, $$\frac{2(2b^2)}{b} = 4\sqrt{3} \implies 4b = 4\sqrt{3} \implies b = \sqrt{3} \implies b^2 = 3$$. Then $$a^2 = 6$$.

Point $$(\alpha, 6)$$ lies on $$H \implies -\frac{\alpha^2}{6} + \frac{36}{3} = 1 \implies \frac{\alpha^2}{6} = 11 \implies \alpha^2 = 66$$.

Product of focal distances for a vertical hyperbola is $$\beta = (ey_1)^2 - b^2 = (3)(36) - 3 = 108 - 3 = 105$$.

Value: $$\alpha^2 + \beta = 66 + 105 = \mathbf{171}$$