Let m 1 , m 2 be the slopes of two adjacent sides of a square of side a such that a 2 + 11 a + 3   m 1…

Given $m_1 \& m_2$ are slopes of two adjacent sides of square then, $m_1{m}_2=-1$

Also given $a^2+11a+3\left(m_1^2+m_2^2\right)=220$

 $\Rightarrow a^2+11a+3\left(m_1^2+\frac{1}{m_1^2}\right)=220$

Now on plotting the diagram we get,

Now given equation of $\mathrm{AC}$,

$\left(\cos \alpha -\sin \alpha \right)x+\left(\sin \alpha +\cos \alpha \right)y=10$

Now by property of square we know that diagonals are perpendicular to each other 

So equation of $BD$ will be $\left(\sin \alpha +\cos \alpha \right)\mathrm{x}+\left(\sin \alpha -\cos \alpha \right)y+\lambda =0$ now is passes through $D\left(10\left(\cos \alpha -\sin \alpha \right),10\left(\sin \alpha -\cos \alpha \right)\right)$,

So, equation of  $BD=\left(\sin \alpha +\cos \alpha \right)\mathrm{x}+\left(\sin \alpha -\cos \alpha \right)y=0$

Now slope of $AB \& AD$ will be given by angle bisector equation of $AC \& BD$

$\frac{\left(\cos \alpha -\sin \alpha \right)x+\left(\sin \alpha +\cos \alpha \right)y-10}{\sqrt{2}}=\pm \frac{\left(\sin \alpha +\cos \alpha \right)\mathrm{x}+\left(\sin \alpha -\cos \alpha \right)y}{\sqrt{2}}$

Taking positive sign we get,

$-2\sin \alpha x+2\cos \alpha y-10=0$, so slope $m_1=\tan \alpha$

Now taking negative sign we get,

$2\cos \alpha x+2\sin \alpha y-10=0$, so slope $m_2=-\text{cot}\alpha$

Now finding $a$ by using distance of point from line 

Now we can see distance of $D$ from $AC$ will be $\frac{a}{\sqrt{2}}$,

So, $\frac{10{\left(\cos \alpha -\sin \alpha \right)}^2+10{\left(\cos \alpha -\sin \alpha \right)}^2-10}{\sqrt{2}}=\frac{a}{\sqrt{2}}$

$\Rightarrow a=10$

Now putting the value of $a, m_1 \& m_2$ in $a^2+11a+3\left(m_1^2+\frac{1}{m_{1}2}\right)=220$ we get,

$\Rightarrow 100+110+3\left({\tan }^2\alpha +{\cot }^2\alpha \right)=220$

$\Rightarrow 210+3\left({\tan }^2\alpha +{\cot }^2\alpha \right)=220$

$\Rightarrow 3\left({\tan }^2\alpha +{\cot }^2\alpha \right)=10$

$\Rightarrow \left({\tan }^2\alpha +{\cot }^2\alpha \right)=\frac{10}{3}$

So this is only possible when ${\tan }^2\alpha =3 \& {\text{cot}}^2\alpha =\frac{1}{3}$ or vice versa

So, ${\tan }^2\alpha =3\Rightarrow \alpha =\frac{\pi }{3}$

So, $72\left({\sin }^4\alpha +{\cos }^4\alpha \right)+a^2-3a+13$

$=72\left(\frac{9}{16}+\frac{1}{16}\right)+100-30+13$

$=72\left(\frac{5}{8}\right)+83=45+83=128$