Let $$C$$ be the circle of minimum area touching the parabola $$y = 6 - x^2$$ and the lines $$y = \sqrt{3}|x|$$. Then, which one of the following points lies on the circle $$C$$?

By the symmetry of the parabola $$y = 6 - x^2$$ and the lines $$y = \sqrt{3}|x|$$ about the $$y$$-axis, the circle of minimum area touching both must be centred on the $$y$$-axis at $$(0, k)$$ with radius $$r$$.

The perpendicular distance from $$(0, k)$$ to the line $$\sqrt{3}\,x - y = 0$$ is $$\dfrac{k}{\sqrt{3+1}} = \dfrac{k}{2}$$. For the circle to touch this line, we need $$r = k/2$$, so $$k = 2r$$.

The lines $$y = \sqrt{3}|x|$$ and the parabola $$y = 6 - x^2$$ intersect where $$6 - x^2 = \sqrt{3}\,x$$, giving $$x^2 + \sqrt{3}\,x - 6 = 0$$, so $$x = \sqrt{3}$$ (positive root) and the intersection point is $$(\sqrt{3}, 3)$$.

For the minimum-area circle, we require it to pass through these intersection points $$(\pm\sqrt{3}, 3)$$. Substituting $$(\sqrt{3}, 3)$$ into the circle equation $$x^2 + (y - 2r)^2 = r^2$$: $$3 + (3 - 2r)^2 = r^2$$, which gives $$3 + 9 - 12r + 4r^2 = r^2$$, i.e., $$3r^2 - 12r + 12 = 0$$, so $$r^2 - 4r + 4 = 0$$, yielding $$(r - 2)^2 = 0$$, hence $$r = 2$$.

The circle has centre $$(0, 4)$$ and radius $$2$$: $$x^2 + (y - 4)^2 = 4$$. Checking option (D), the point $$(2, 4)$$: $$4 + 0 = 4$$ ✓. The answer is $$\boxed{(2, 4)}$$, i.e., Option (D).