To solve the given problem, we need to verify the conditions under which the points \(((x, y) = (3 \tan (\theta + \frac{\pi}{3}), 2 \tan (\theta + \frac{\pi}{6})))\) satisfy the equation \((xy + ax + by + \gamma = 0)\) for \((\theta \in [-\frac{\pi}{3}, 0])\).

Let's start by calculating the values for \(x\) and \(y\):

For the \(x\)-coordinate: \(x = 3 \tan (\theta + \frac{\pi}{3})\).

Using the tangent addition formula, \(\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}\), we get:

\(\tan(\theta + \frac{\pi}{3}) = \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\).

Hence, \(x = 3 \left(\frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\right)\).

For the \(y\)-coordinate: \(y = 2 \tan (\theta + \frac{\pi}{6})\).

Using the tangent addition formula again, we have:

\(\tan(\theta + \frac{\pi}{6}) = \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \frac{1}{\sqrt{3}}}\).

Hence, \(y = 2 \left(\frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \frac{1}{\sqrt{3}}}\right)\).

Now, substitute these into the equation \((xy + ax + by + \gamma = 0)\):

\(\left(3 \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\right) \left(2 \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \frac{1}{\sqrt{3}}}\right) + a \left(3 \frac{\tan \theta + \sqrt{3}}{1 - \tan \theta \sqrt{3}}\right) + b \left(2 \frac{\tan \theta + \frac{1}{\sqrt{3}}}{1 - \tan \theta \frac{1}{\sqrt{3}}}\right) + \gamma = 0\)

Simplify it to get the condition that holds for all \((\theta)\) in the specified range.

To compute \(a^2 + b^2 + \gamma^2\), let's assume some instance or particular cases to deduce the actual values:

After analyzing these conditions through simplification and employing trials for specific values of \((\theta)\), we independently solve the system to deduce values for \(a, b, \text{ and } \gamma\) and compute \((a^2 + b^2 + \gamma^2)\) which gives:

\(a^2 + b^2 + \gamma^2 = 75\).

Therefore, the correct option is 75.