Circle

If the equation of the circle passing through the points of intersection of the circles $x^2-2 x+y^2-4 y-4=0$, $x^2+2 x+y^2+4 y-4=0$ and the point $(3,3)$ is given by $x^2+y^2+\alpha x+\beta y+\gamma=0$, then $3(\alpha+\beta+\gamma)=$