Let $$m_1, m_2$$ be the slopes of two adjacent sides of a square of side a such that $$a^2 + 11a + 3(m_1^2 + m_2^2) = 220$$. If one vertex of the square is $$10(\cos\alpha - \sin\alpha, \sin\alpha + \cos\alpha)$$, where $$\alpha \in (0, \frac{\pi}{2})$$ and the equation of one diagonal is $$(\cos\alpha - \sin\alpha)x + (\sin\alpha + \cos\alpha)y = 10$$, then $$72(\sin^4\alpha + \cos^4\alpha) + a^2 - 3a + 13$$ is equal to

We have a square of side $$a$$ with adjacent side slopes $$m_1, m_2$$ satisfying $$a^2 + 11a + 3(m_1^2 + m_2^2) = 220$$. One vertex is $$P = 10(\cos\alpha - \sin\alpha,\; \sin\alpha + \cos\alpha)$$ and one diagonal has equation $$(\cos\alpha - \sin\alpha)x + (\sin\alpha + \cos\alpha)y = 10$$.

We note that vertex $$P$$ lies on the diagonal line. Substituting $$P$$ into the diagonal equation: $$(\cos\alpha - \sin\alpha) \cdot 10(\cos\alpha - \sin\alpha) + (\sin\alpha + \cos\alpha) \cdot 10(\sin\alpha + \cos\alpha) = 10(\cos\alpha - \sin\alpha)^2 + 10(\sin\alpha + \cos\alpha)^2 = 10[(\cos^2\alpha - 2\sin\alpha\cos\alpha + \sin^2\alpha) + (\sin^2\alpha + 2\sin\alpha\cos\alpha + \cos^2\alpha)] = 10[1 + 1] = 20 \neq 10$$.

So $$P$$ does not lie on this diagonal. Since a square has two diagonals, $$P$$ is a vertex, and one diagonal is given, $$P$$ must be a vertex not on this diagonal. The distance from $$P$$ to the diagonal equals half the other diagonal, which is $$\dfrac{a}{\sqrt{2}}$$ (since both diagonals of a square have length $$a\sqrt{2}$$ and the distance from a vertex to the opposite diagonal is $$\dfrac{a\sqrt{2}}{2}$$).

The distance from $$P$$ to the line $$(\cos\alpha - \sin\alpha)x + (\sin\alpha + \cos\alpha)y = 10$$ is: $$d = \dfrac{|20 - 10|}{\sqrt{(\cos\alpha - \sin\alpha)^2 + (\sin\alpha + \cos\alpha)^2}} = \dfrac{10}{\sqrt{2}}$$

For a square, the distance from any vertex to the opposite diagonal is $$\dfrac{a\sqrt{2}}{2}$$. So $$\dfrac{10}{\sqrt{2}} = \dfrac{a\sqrt{2}}{2}$$, giving $$\dfrac{10}{\sqrt{2}} = \dfrac{a}{\sqrt{2}}$$, so $$a = 10$$.

Now, since adjacent sides of a square are perpendicular, $$m_1 \cdot m_2 = -1$$. The diagonal direction is along $$(-(\sin\alpha + \cos\alpha),\; \cos\alpha - \sin\alpha)$$ (normal to the line). The diagonal slope is $$\dfrac{\cos\alpha - \sin\alpha}{-(\sin\alpha + \cos\alpha)}$$. The sides of the square make 45° with the diagonal, so the slopes of the sides are related to the diagonal slope.

The diagonal direction vector is proportional to $$(\sin\alpha + \cos\alpha,\; -(\cos\alpha - \sin\alpha)) = (\sin\alpha + \cos\alpha,\; \sin\alpha - \cos\alpha)$$. The slope of this diagonal is $$\dfrac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}$$.

For the sides making $$45°$$ with this diagonal, using the formula $$\tan 45° = \left|\dfrac{m - m_d}{1 + m \cdot m_d}\right|$$, and since both sides are perpendicular to each other, one slope comes from $$+45°$$ and the other from $$-45°$$ rotation.

Let $$m_d = \dfrac{\sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha}$$. Then $$m_1 = \dfrac{m_d + 1}{1 - m_d}$$ and $$m_2 = \dfrac{m_d - 1}{1 + m_d}$$.

Now $$m_1 = \dfrac{\sin\alpha - \cos\alpha + \sin\alpha + \cos\alpha}{\sin\alpha + \cos\alpha - \sin\alpha + \cos\alpha} = \dfrac{2\sin\alpha}{2\cos\alpha} = \tan\alpha$$.

Similarly, $$m_2 = \dfrac{\sin\alpha - \cos\alpha - \sin\alpha - \cos\alpha}{\sin\alpha + \cos\alpha + \sin\alpha - \cos\alpha} = \dfrac{-2\cos\alpha}{2\sin\alpha} = -\cot\alpha$$.

So $$m_1^2 + m_2^2 = \tan^2\alpha + \cot^2\alpha$$.

Substituting into $$a^2 + 11a + 3(m_1^2 + m_2^2) = 220$$ with $$a = 10$$: $$100 + 110 + 3(\tan^2\alpha + \cot^2\alpha) = 220$$, so $$3(\tan^2\alpha + \cot^2\alpha) = 10$$.

Now we need $$72(\sin^4\alpha + \cos^4\alpha) + a^2 - 3a + 13$$. With $$a = 10$$: $$a^2 - 3a + 13 = 100 - 30 + 13 = 83$$.

We have $$\tan^2\alpha + \cot^2\alpha = \dfrac{\sin^4\alpha + \cos^4\alpha}{\sin^2\alpha\cos^2\alpha} = \dfrac{10}{3}$$.

Let $$p = \sin^2\alpha + \cos^2\alpha = 1$$ and $$q = \sin^2\alpha\cos^2\alpha$$. Then $$\sin^4\alpha + \cos^4\alpha = 1 - 2q$$. The condition becomes $$\dfrac{1-2q}{q} = \dfrac{10}{3}$$, so $$3 - 6q = 10q$$, giving $$16q = 3$$, so $$q = \dfrac{3}{16}$$. Therefore $$\sin^4\alpha + \cos^4\alpha = 1 - \dfrac{6}{16} = 1 - \dfrac{3}{8} = \dfrac{5}{8}$$.

Finally, $$72 \cdot \dfrac{5}{8} + 83 = 45 + 83 = 128$$.

Hence, the correct answer is Option B: $$128$$.