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, $C\left(\frac{12+5}{2},\frac{7+8}{2}\right)$

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

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

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

$=1+16=17$