Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle $$\alpha$$ with the positive x-axis and the equations of its diagonals are $$(\sqrt{3}+1)x + (\sqrt{3}-1)y = 0$$ and $$(\sqrt{3}-1)x - (\sqrt{3}+1)y + 8\sqrt{3} = 0$$. Then $$a^2$$ is equal to

The vertices of the square are $$O(0,0),\;A,\;B,\;C$$. Because $$OABC$$ is a square, the two diagonals are $$OC$$ and $$AB$$. The line that passes through the origin must be $$OC$$, and the other line must be $$AB$$.

Equation of diagonal $$OC$$ (through $$O$$): $$(\sqrt{3}+1)x + (\sqrt{3}-1)y = 0 \quad -(1)$$

Equation of diagonal $$AB$$: $$(\sqrt{3}-1)x - (\sqrt{3}+1)y + 8\sqrt{3} = 0 \quad -(2)$$

The diagonals of a square intersect at their common midpoint $$M$$. Thus, the point of intersection of $$(1)$$ and $$(2)$$ gives the coordinates of $$M$$.

From $$(1)$$: $$(\sqrt{3}+1)x = -(\sqrt{3}-1)y \;\;\Rightarrow\;\; x = -\frac{\sqrt{3}-1}{\sqrt{3}+1}y \quad -(3)$$

Put $$(3)$$ into $$(2)$$:

$$(\sqrt{3}-1)\left(-\frac{\sqrt{3}-1}{\sqrt{3}+1}y\right) - (\sqrt{3}+1)y + 8\sqrt{3} = 0$$

Simplify the y-coefficients first.
Define $$r = \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$$. Then the coefficient of $$y$$ becomes $$-(\sqrt{3}-1)r - (\sqrt{3}+1).$$

Compute $$r$$ by rationalising the denominator:
$$r = \frac{\sqrt{3}-1}{\sqrt{3}+1}\cdot\frac{\sqrt{3}-1}{\sqrt{3}-1} = \frac{(\sqrt{3}-1)^2}{2} = 2-\sqrt{3}.$$

Now $$-(\sqrt{3}-1)r - (\sqrt{3}+1) = -(\sqrt{3}-1)(2-\sqrt{3}) - (\sqrt{3}+1) = -\frac{8}{\sqrt{3}+1}.$$

Hence $$-\frac{8}{\sqrt{3}+1}\,y + 8\sqrt{3} = 0 \;\;\Longrightarrow\;\; y = \sqrt{3}(\sqrt{3}+1) = 3+\sqrt{3}.$$

Using $$(3)$$ with $$r = 2-\sqrt{3}$$:
$$x = -ry = -(2-\sqrt{3})(3+\sqrt{3}) = -\bigl(3-\sqrt{3}\bigr) = \sqrt{3}-3.$$

Therefore $$M\bigl(\,\sqrt{3}-3,\; 3+\sqrt{3}\bigr).$$

Distance $$OM$$: $$OM^2 = (\sqrt{3}-3)^2 + (3+\sqrt{3})^2 = (12-6\sqrt{3}) + (12+6\sqrt{3}) = 24.$$

In a square, the midpoint $$M$$ divides diagonal $$OC$$ in the ratio $$1:1$$, so $$OM = \frac{OC}{2}.$$ For a square of side $$a$$, the diagonal length is $$OC = a\sqrt{2}$$, hence $$OM = \frac{a\sqrt{2}}{2} \quad\Longrightarrow\quad OM^2 = \frac{a^2}{2}.$$

Equate the two expressions for $$OM^2$$:
$$\frac{a^2}{2} = 24 \;\;\Longrightarrow\;\; a^2 = 48.$$

Thus, $$a^2 = 48$$, which corresponds to Option A.