Let (5, (a)/(4)) , be the circumcenter of a triangle with vertices...

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

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

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

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

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

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

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

Using \left(i\right), \left(ii\right)\text{and}\left(iii\right)$,</p><p>$⇒ A{C}^{2} + A{B}^{2} = B{C}^{2}$</p><p>So,$△ ABC$is right angled at$A$.</p><p>So, circumcentre will be mid-point of$BC$.</p><p>$⇒ S ≡ \left(\dfrac{a + \dfrac{a}{4}}{2}, \dfrac{6 - 2}{2}\right)$</p><p>$⇒ S ≡ \left(\dfrac{5a}{8}, 2\right) = \left(5, \dfrac{a}{4}\right)$</p><p>$⇒ a = 8$</p><p>$⇒ BC = \sqrt{\dfrac{9 \times 64}{16} + 64}$</p><p>$⇒ BC = \sqrt{100} = 10$</p><p>$⇒ AB = 8$</p><p>$⇒ AC = \sqrt{\dfrac{9 \times 64}{16}}$</p><p>$⇒ AC = 6$</p><p>Now, circumradius is given by,</p><p>$R = \dfrac{BC}{2} = 5 ⇒ \alpha = 5$</p><p>Now, area of triangle will be$\beta = \dfrac{1}{2} \times 6 \times 8 = 24 and perimeter will be $\gamma = 6 + 8 + 10 = 24$

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

$⇒ \alpha + \beta + \gamma = 53$