Let ${\mathrm{m}}_1,{\mathrm{m}}_2$ be the slopes of two adjacent sides of a square of side a such that ${\mathrm{a}}^2+11\mathrm{a}+3\left({\mathrm{m}}_1^2+{\mathrm{m}}_2^2\right)=220$. If one vertex of the square is $\left(10\right(\cos \mathrm{\alpha }-\sin \mathrm{\alpha })$ $10(\sin \mathrm{\alpha }+\cos \mathrm{\alpha }))$, where $\mathrm{\alpha }\in \left(0,\frac{\mathrm{\pi }}{2}\right)$ and the equation of one diagonal is $(\cos \mathrm{\alpha }-\sin \mathrm{\alpha })\mathrm{x}+(\sin \mathrm{\alpha }+\cos \mathrm{\alpha })\mathrm{y}=10$ then $72\left({\sin }^4\mathrm{\alpha }+{\cos }^4\mathrm{\alpha }\right)+{\mathrm{a}}^2-3\mathrm{a}+13$ is equal to:

$\begin{array}{l}\mathrm{AC}=(\cos \mathrm{\alpha }-\sin \mathrm{\alpha })+(\sin \mathrm{\alpha }+\cos \mathrm{\alpha })\mathrm{y}=10\\ \mathrm{BD}=(\sin \mathrm{\alpha }-\cos \mathrm{\alpha })\mathrm{x}+(\sin \mathrm{\alpha }-\cos \mathrm{\alpha })\mathrm{y}=0\\ \\ \mathrm{point} \mathrm{of} \mathrm{intersection} \mathrm{of} \mathrm{diagonals} \mathrm{is}\left(5\right(\cos \mathrm{\alpha }-\sin \mathrm{\alpha }),5(\cos \mathrm{\alpha }+\sin \mathrm{\alpha }\left)\right)\\ \mathrm{one} \mathrm{vertex} \mathrm{of} \mathrm{the} \mathrm{square} \mathrm{is} \left(10\right(\cos \mathrm{\alpha }-\sin \mathrm{\alpha }),10(\sin \mathrm{\alpha }-\cos \mathrm{\alpha }\left)\right)\\ \mathrm{Length} \mathrm{of} \mathrm{the} \mathrm{diagonal} \mathrm{is} 10\sqrt{2}\Rightarrow \mathrm{length} \mathrm{of} \mathrm{side} \mathrm{a}=10\end{array}$

$\begin{array}{r}{a}^2+11a+3\left(m_1^2+m_2^2\right)=220\\ m_1^2+m_2^2=\frac{220-100-110}{3}=\frac{10}{3}\\ \text{ and }{m}_1{m}_2=-1\\ \text{ Slopes of the sides are tanα and - cotα }\\ \begin{array}{r}{\tan }^2\alpha =3\text{ or }\frac{1}{3}\\ 72\left({\sin }^4\alpha +{\cos }^4\alpha \right)+a^2-3a+13\\ =72\cdot \frac{{\tan }^4\alpha +1}{{\left(1+{\tan }^2\alpha \right)}^2}+a^2-3a+13=128\end{array}\end{array}$