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:

Step 1: Define x and y in terms of new variables

Let's simplify the expressions for $x$ and $y$ by introducing new variables. We define $A = \theta + \frac{\pi}{3}$ and $B = \theta + \frac{\pi}{6}$. This makes the expressions for $x$ and $y$ more manageable.

Step 2: Find the relationship between A and B

Now, let's find the relationship between $A$ and $B$. Subtracting $B$ from $A$, we get $(\theta + \frac{\pi}{3}) - (\theta + \frac{\pi}{6}) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{2\pi - \pi}{6} = \frac{\pi}{6}$. This constant difference will be useful.

Step 3: Apply the tangent subtraction formula

We know the value of $A - B$, so we can use the tangent subtraction formula. The formula states that $\tan(A - B)$ is equal to $(\tan A - \tan B)$ divided by $(1 + \tan A \tan B)$.

Step 4: Substitute and rearrange the equation

Substitute $\tan(A - B) = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$, $\tan A = \frac{x}{3}$, and $\tan B = \frac{y}{2}$ into the tangent formula. After cross-multiplication and rearrangement, we get $xy - 2x + 3y - 18 = 0$. This is the equation of the locus.

Step 5: Compare with the given equation to find alpha, beta, gamma

The given equation is $xy + \alpha x + \beta y + \gamma = 0$. By comparing this with our derived equation $xy - 2x + 3y - 18 = 0$, we can identify the values of $\alpha$, $\beta$, and $\gamma$.

Step 6: Calculate the final expression

Finally, we need to calculate $\alpha^2 + \beta^2 + \gamma^2$. Substitute the values we found: $(-2)^2 + (3)^2 + (-18)^2 = 4 + 9 + 324 = 337$.

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