Let $\left(5,\dfrac{9}{4}\right)$ be the circumcenter of a triangle with vertices $A(a,-2)$, $B(a,6)$ and $C\!\left(\dfrac{a}{4},-2\right)$. Let $\alpha$ denote the circumradius, $\beta$ denote the area and $\gamma$ denote the perimeter of the triangle. Then $\alpha+\beta+\gamma$ is: