$$\begin{aligned} &\text { Let } \phi=\theta+\frac{\pi}{3} \Rightarrow \theta=\phi-\frac{\pi}{6}\\ &\begin{aligned} & x=3 \tan \left(\theta+\frac{\pi}{3}\right)=3 \tan \left(\theta+\frac{\pi}{6}\right) \\ & y=2 \tan \phi \\ & \tan \left(\phi+\frac{\pi}{6}\right)=\frac{\tan \phi+\frac{1}{\sqrt{3}}}{1-\tan \phi \cdot \frac{1}{\sqrt{3}}} \\ & \frac{x}{3}=\frac{\frac{y}{2}+\frac{1}{\sqrt{3}}}{1-\frac{y}{2} \cdot \frac{1}{\sqrt{3}}} \end{aligned} \end{aligned}$$

$$\begin{aligned} & \Rightarrow \quad x=\frac{3(y \sqrt{3}+2)}{2 \sqrt{3}-y} \\ & x y+\alpha x+\beta y+r=0 \\ & 3\left(\frac{y \sqrt{3}+2}{2 \sqrt{3}-y}\right)+\alpha\left(3 \frac{(y \sqrt{3}+2)}{(2 \sqrt{3}-y)}\right)+\beta y+r=0 \\ & =(3 \sqrt{3}-\beta) y^2+(6+3 \sqrt{3} \alpha+2 \sqrt{3} \beta-y) y \\ & \quad+(6 \alpha+2 \sqrt{3} y)=0 \end{aligned}$$

For this identity to hold for all $\theta$, coefficients must be 0

$$\begin{aligned} & \therefore \quad \beta=3 \sqrt{3} \\ & \gamma=-\alpha \sqrt{3} \\ & 6+3 \sqrt{3} \alpha+(2 \sqrt{3})(3 \sqrt{3})+\alpha \sqrt{3}=0 \\ & \Rightarrow \alpha=-2 \sqrt{3} \\ & \Rightarrow \beta=6 \\ & \alpha^2+\beta^2+\gamma^2=75 \end{aligned}$$