Let (5, a/4) be the circumcenter of a triangle with vertices A (a,-2), B (a, 6) and C(a/4,-2). Let α denote the circumradius, β denote the area and γ…

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