Let $\mathrm{C}(\mathrm{\alpha },\mathrm{\beta })$ be the circumcenter of the triangle formed by the lines 4x + 3y = 69, 4y – 3x = 17 and x + 7y = 61. Then  ${(\mathrm{\alpha }-\mathrm{\beta })}^2+\mathrm{\alpha }+\mathrm{\beta }$ is equal to 

$\begin{array}{l}{\mathrm{M}}_1\perp {\mathrm{M}}_2\\ \therefore \mathrm{C}\text{ is mid point of hypotenuse }\\ 4\mathrm{x}+3\mathrm{y}=69\Rightarrow \mathrm{y}=7,\mathrm{x}=12\\ \mathrm{x}+7\mathrm{y}=61\\ 4\mathrm{y}-3\mathrm{x}=17\Rightarrow \mathrm{y}=8,\mathrm{x}=5\\ \mathrm{x}+7\mathrm{y}=61\\ \therefore P(12, 7), Q(5, 8)\end{array}$
$\begin{array}{l}C\left(\frac{12+5}{2},\frac{7+8}{2}\right)\\ C\left(\frac{17}{2},\frac{15}{2}\right)=(\alpha ,\beta )\\ (\alpha -\beta {)}^2+\alpha +\beta \\ ={\left(\frac{17}{2}-\frac{15}{2}\right)}^2+\frac{17}{2}+\frac{15}{2}\\ =1+16=17\end{array}$