If for $$\theta \in \left[-\frac{\pi}{3}, 0\right]$$, the points $$(x, y) = \left(3\tan\left(\theta + \frac{\pi}{3}\right), 2\tan\left(\theta + \frac{\pi}{6}\right)\right)$$ lie on $$xy + \alpha x + \beta y + \gamma = 0$$, then $$\alpha^2 + \beta^2 + \gamma^2$$ is equal to :

We have $$x = 3\tan\left(\theta + \frac{\pi}{3}\right)$$ and $$y = 2\tan\left(\theta + \frac{\pi}{6}\right)$$ for $$\theta \in \left[-\frac{\pi}{3}, 0\right]$$.

Let $$\alpha = \theta + \frac{\pi}{3}$$ and $$\beta = \theta + \frac{\pi}{6}$$, so $$\alpha - \beta = \frac{\pi}{6}$$, $$\tan\alpha = \frac{x}{3}$$, and $$\tan\beta = \frac{y}{2}$$.

Using the tangent subtraction formula:

$$\tan(\alpha - \beta) = \frac{\tan\alpha - \tan\beta}{1 + \tan\alpha \tan\beta} = \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$$

Substituting: $$\frac{\frac{x}{3} - \frac{y}{2}}{1 + \frac{xy}{6}} = \frac{1}{\sqrt{3}}$$

Cross-multiplying: $$\sqrt{3}\left(\frac{x}{3} - \frac{y}{2}\right) = 1 + \frac{xy}{6}$$

Multiplying through by 6: $$2\sqrt{3}x - 3\sqrt{3}y = 6 + xy$$

Rearranging: $$xy - 2\sqrt{3}x + 3\sqrt{3}y + 6 = 0$$

Comparing with $$xy + \alpha x + \beta y + \gamma = 0$$, we get $$\alpha = -2\sqrt{3}$$, $$\beta = 3\sqrt{3}$$, $$\gamma = 6$$.

Therefore, $$\alpha^2 + \beta^2 + \gamma^2 = 12 + 27 + 36 = 75$$.

Hence, the correct answer is Option D.