An equilateral triangle OAB is inscribed in the parabola $$y^{2} = 4x$$ with the vertex O at the vertex of the parabola. Then the minimum distance of the circle having AB as a diameter from the origin is

We need to find the minimum distance from the origin to the circle having AB as diameter, where equilateral triangle OAB is inscribed in the parabola $$y^2 = 4x$$ with O at the vertex.

Since the parabola $$y^2 = 4x$$ has vertex at the origin O(0,0) and parameter $$a = 1$$, any point on it can be written as $$(t^2, 2t)$$. By symmetry about the x-axis, let $$A = (t^2, 2t)$$ and $$B = (t^2, -2t)$$ for some $$t > 0$$.

Next, since $$AB = |2t - (-2t)| = 4t$$ and $$OA = \sqrt{t^4 + 4t^2} = t\sqrt{t^2 + 4}$$, imposing the equilateral condition $$OA = AB$$ gives $$t\sqrt{t^2 + 4} = 4t$$ which simplifies to $$\sqrt{t^2 + 4} = 4 \implies t^2 + 4 = 16 \implies t^2 = 12 \implies t = 2\sqrt{3}$$.

Using this value yields $$A = (12, 4\sqrt{3})$$ and $$B = (12, -4\sqrt{3})$$. The midpoint of $$AB$$, which is the center of the circle having $$AB$$ as diameter, is $$M = (12, 0)$$, and its radius is $$\frac{AB}{2} = \frac{4 \times 2\sqrt{3}}{2} = 4\sqrt{3}$$.

Since the distance from the origin to the center is $$OM = \sqrt{12^2 + 0^2} = 12$$, the minimum distance from the origin to the circle is $$OM - r = 12 - 4\sqrt{3} = 4(3 - \sqrt{3})$$.

The correct answer is Option 4: $$4(3 - \sqrt{3})$$.