1. Use the slopes of adjacent sides of a square

For a square, adjacent sides are perpendicular, so if their slopes are $m_1,m_2$, then $m_1m_2=-1.$ Hence one slope is $m$ and the other is $-\frac1m$.

So, $m_1^2+m_2^2=m^2+\frac{1}{m^2}\ge 2,$ with equality when $m^2=1$.

Given $a^2+11a+3(m_1^2+m_2^2)=220,$ we get $a^2+11a+3\left(m^2+\frac1{m^2}\right)=220.$ Since $m^2+\frac1{m^2}\ge 2$, $a^2+11a+6\le 220,$ so $a^2+11a\le 214.$

Now check integer-looking possibility from options context. If $m^2+\frac1{m^2}=2$, then $a^2+11a+6=220$ $a^2+11a-214=0$ $(a-11)(a+22)=0.$ Since side length is positive, $a=11.$ Thus necessarily $m_1^2+m_2^2=2.$

2. Interpret the given vertex and diagonal

Let $P=(10(\cos\alpha-\sin\alpha),\;10(\sin\alpha+\cos\alpha)).$

The given diagonal is $(\cos\alpha-\sin\alpha)x+(\sin\alpha+\cos\alpha)y=10.$

Substitute $P$ into the left side: \begin{align*} &(\cos\alpha-\sin\alpha)\cdot 10(\cos\alpha-\sin\alpha) +(\sin\alpha+\cos\alpha)\cdot 10(\sin\alpha+\cos\alpha)\ &=10\left[(\cos\alpha-\sin\alpha)^2+(\sin\alpha+\cos\alpha)^2\right]. \end{align*} Now, \begin{align*} (\cos\alpha-\sin\alpha)^2+(\sin\alpha+\cos\alpha)^2 &=(\cos^2\alpha+\sin^2\alpha-2\sin\alpha\cos\alpha)\ &\quad +(\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha)\ &=2. \end{align*} So LHS $=20\ne 10$. Therefore this vertex is not on that diagonal.

In a square, a vertex not lying on a diagonal must be an endpoint of the other diagonal's opposite pair relation. The center is on both diagonals.

3. Find the center of the square from the given diagonal

The diagonal has equation $(\cos\alpha-\sin\alpha)x+(\sin\alpha+\cos\alpha)y=10.$ Its normal vector is $\bigl(\cos\alpha-\sin\alpha,\;\sin\alpha+\cos\alpha\bigr).$ Notice that the given vertex is $P=10\bigl(\cos\alpha-\sin\alpha,\;\sin\alpha+\cos\alpha\bigr),$ so $P$ lies along the normal direction.

For a square, the center lies on the line joining a vertex to the midpoint and also on the diagonal. Thus the foot of perpendicular from origin structure strongly suggests using distance from $P$ to the diagonal as half-diagonal.

Distance of point $P(x_1,y_1)$ from the line $Ax+By-10=0$ is $d=\frac{|Ax_1+By_1-10|}{\sqrt{A^2+B^2}}.$ Here $A=\cos\alpha-\sin\alpha,\quad B=\sin\alpha+\cos\alpha.$ We already found $Ax_1+By_1=20,$ so $d=\frac{|20-10|}{\sqrt{A^2+B^2}}=\frac{10}{\sqrt{2}},$ because $A^2+B^2=2.$ Hence $d=5\sqrt2.$

This distance equals the perpendicular distance from a vertex to a diagonal not containing it, which in a square is exactly half the diagonal length: $\frac{a\sqrt2}{2}=\frac{a}{\sqrt2}.$ So $\frac{a}{\sqrt2}=5\sqrt2\implies a=10.$

But this contradicts the earlier value $a=11$. So let us correct the interpretation: in a square, distance from a vertex to the opposite diagonal is actually $\frac{2\cdot (\text{area of one of the triangles})}{\text{diagonal length}}=\frac{a^2}{a\sqrt2}=\frac{a}{\sqrt2},$ which again gives $a=10$. Therefore the only way to reconcile the condition involving slopes is that we should determine $a$ directly from geometry and then use the slope condition to infer orientation.

Substitute $a=10$ in $a^2+11a+3(m_1^2+m_2^2)=220$ $100+110+3(m_1^2+m_2^2)=220$ $3(m_1^2+m_2^2)=10,$ which is impossible for perpendicular slopes since $m_1^2+m_2^2\ge 2$ gives RHS at least $6$, actually possible; but exact value $10/3>2$ is possible. So $a=10$ is consistent.

4. Find $\sin^4\alpha+\cos^4\alpha$

Use identity

Now the diagonal has normal vector $\left(\cos\alpha-\sin\alpha,\;\sin\alpha+\cos\alpha\right),$ whose slope is

This simplifies to

Since the square sides are at $\pm 45^\circ$ to diagonal directions and from the earlier slope relation the adjacent side slopes give symmetric value, we get the square is oriented so that $m_1^2+m_2^2=\frac{10}{3}.$ For perpendicular slopes $m,-1/m$,

This gives

Thus side slopes are $\pm\sqrt3,\mp\frac1{\sqrt3}$, i.e. directions $30^\circ$ and $120^\circ$, so diagonals make angles $75^\circ$ and $-15^\circ$ with x-axis. Comparing with diagonal equation gives $\frac\pi4-\alpha=15^\circ=\frac\pi{12},$ so $\alpha=\frac\pi6.$

Therefore