If for θ ∈[-(π)/3, 0], the points (x, y)=(3 tan (θ+(π)/3), 2 tan (θ+(π)/6)) lie on x y+α x+β y+γ=0, then α^2+β^2+γ^2 is equal to

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