Let m _ 1 , m _ 2 be the slopes of two adjacent sides of a square...

Given ${m}_{1} \text{\text{ and } }$ {m}_{2} are slopes of two adjacent sides of square then, ${m}_{1}{m}_{2} = - 1$

Also given ${a}^{2} + 11a + 3\left({m}_{1}^{2} + {m}_{2}^{2}\right) = 220$

$⇒ {a}^{2} + 11a + 3\left({m}_{1}^{2} + 1{m}_{1}^{2}\right) = 220$

Now on plotting the diagram we get,

Now given equation of $AC$,

\left(cos\alpha - sin\alpha\right)x + \left(sin\alpha + cos\alpha\right)y = 10$</p><p>Now by property of square we know that diagonals are perpendicular to each other </p><p>So equation of$BD$will be \left(sin\alpha + cos\alpha\right)x + \left(sin\alpha - cos\alpha\right)y + \lambda = 0$ now is passes through $D\left(10\left(cos\alpha - sin\alpha\right), 10\left(sin\alpha - cos\alpha\right)\right)$,

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

Now slope of $AB \text{ and } AD$ will be given by angle bisector equation of $AC \text{ and } BD$

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

Taking positive sign we get,

$- 2sin\alphax + 2cos\alphay - 10 = 0$, so slope ${m}_{1} = tan\alpha$

Now taking negative sign we get,

$2cos\alphax + 2sin\alphay - 10 = 0$, 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 $\dfrac{a}{\sqrt{2}}$,

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

$⇒ a = 10$

Now putting the value of $a, {m}_{1} \text{ and } {m}_{2} in {a}^{2} + 11a + 3\left({m}_{1}^{2} + \dfrac{1}{{m}_{1}2}\right) = 220$ we get,

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

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

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

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

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

So, $\tan^{2}\alpha = 3 ⇒ \alpha = \dfrac{\pi}{3}$

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

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

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