Let C ( α, β ) be the circumcentre of the triangle formed by the...

Given,

$C ( \alpha, \beta )$ be the circumcentre of the triangle formed by the lines $4x + 3y = 69$, $4y - 3x = 17$, and $x + 7y = 61$,

Now plotting the diagram of the above equations and finding the intersection point we get,

Now from diagram we can see that it is a right-angled triangle, so the circumcentre is given by midpoint of the hypotenuse,

Hence, circumcentre will be, $C\left(\dfrac{12 + 5}{2}, \dfrac{7 + 8}{2}\right)$

$⇒ C\left(\dfrac{17}{2}, \dfrac{15}{2}\right) = ( \alpha, \beta )$

Hence, $( \alpha - \beta{)}^{2} + \alpha + \beta$

$= {\left(\dfrac{17}{2} - \dfrac{15}{2}\right)}^{2} + \dfrac{17}{2} + \dfrac{15}{2}$

$= 1 + 16 = 17$