Given: $A\left(a,-2\right), B\left(a,6\right) \text{and} C\left(\frac{a}{4},-2\right)$

$\Rightarrow A{B}^2={\left(a-a\right)}^2+{\left(6+2\right)}^2$

$\Rightarrow A{B}^2=64 ...\left(i\right)$

$\Rightarrow B{C}^2={\left(a-\frac{a}{4}\right)}^2+{\left(6+2\right)}^2$

$\Rightarrow B{C}^2=\frac{9a^2}{16}+64 ...\left(ii\right)$

$\Rightarrow A{C}^2={\left(a-\frac{a}{4}\right)}^2+{\left(-2+2\right)}^2$

$\Rightarrow A{C}^2=\frac{9a^2}{16} ...\left(iii\right)$

Using $\left(i\right), \left(ii\right) \text{and} \left(iii\right)$,

$\Rightarrow A{C}^2+A{B}^2=B{C}^2$

So, $△ABC$ is right angled at $A$.

So, circumcentre will be mid-point of $BC$.

$\Rightarrow S\equiv \left(\frac{a+{\displaystyle \frac{a}{4}}}{2},\frac{6-2}{2}\right)$

$\Rightarrow S\equiv \left(\frac{5a}{8},2\right)=\left(5,\frac{a}{4}\right)$

$\Rightarrow a=8$

$\Rightarrow BC=\sqrt{\frac{9\times 64}{16}+64}$

$\Rightarrow BC=\sqrt{100}=10$

$\Rightarrow AB=8$

$\Rightarrow AC=\sqrt{\frac{9\times 64}{16}}$

$\Rightarrow AC=6$

Now, circumradius is given by,

$R=\frac{BC}{2}=5\Rightarrow \alpha =5$

Now, area of triangle will be $\beta =\frac{1}{2}\times 6\times 8=24$ and perimeter will be $\gamma =6+8+10=24$

$\Rightarrow \alpha =5, \beta =24, \gamma =24$

$\Rightarrow \alpha +\beta +\gamma =53$