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

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: $$4(3 - \sqrt{3})$$.