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: