Let $$p, q \in \mathbb{R}$$ and $$(1 - \sqrt{3}i)^{200} = 2^{199}(p + iq)$$, $$i = \sqrt{-1}$$. Then, $$p + q + q^2$$ and $$p - q + q^2$$ are roots of the equation.

We need to find $$(1 - \sqrt{3}i)^{200} = 2^{199}(p + iq)$$.

Writing $$1 - \sqrt{3}i$$ in polar form: its modulus is $$|1 - \sqrt{3}i| = \sqrt{1 + 3} = 2$$ and its argument is $$\arg(1 - \sqrt{3}i) = -\frac{\pi}{3}$$, so that $$1 - \sqrt{3}i = 2e^{-i\pi/3}$$.

Raising this to the 200th power gives $$(1 - \sqrt{3}i)^{200} = 2^{200} e^{-i \cdot 200\pi/3}$$. Now, $$\frac{200}{3} = 66\frac{2}{3}$$, so $$\frac{200\pi}{3} = 66\pi + \frac{2\pi}{3}$$ and, since $$e^{-i\cdot 66\pi} = 1$$ (because 66 is even), we have $$e^{-i(66\pi + 2\pi/3)} = e^{-i\cdot 2\pi/3}$$.

Using $$e^{-i \cdot 2\pi/3} = \cos\frac{2\pi}{3} - i\sin\frac{2\pi}{3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}$$, it follows that $$(1 - \sqrt{3}i)^{200} = 2^{200}\Bigl(-\frac{1}{2} - i\frac{\sqrt{3}}{2}\Bigr) = 2^{199}(-1 - i\sqrt{3}).$$ Therefore, $$p = -1$$ and $$q = -\sqrt{3}$$ in the expression $$(1 - \sqrt{3}i)^{200} = 2^{199}(p + iq).$$

Next, we compute $$p + q + q^2 = -1 + (-\sqrt{3}) + 3 = 2 - \sqrt{3}$$ and $$p - q + q^2 = -1 - (-\sqrt{3}) + 3 = 2 + \sqrt{3}$$. These two values will be the roots of the desired quadratic equation.

The sum of the roots is $$(2 - \sqrt{3}) + (2 + \sqrt{3}) = 4$$ and their product is $$(2 - \sqrt{3})(2 + \sqrt{3}) = 4 - 3 = 1$$, so the equation with these roots is $$x^2 - 4x + 1 = 0$$.

The answer is Option 2: $$x^2 - 4x + 1 = 0$$.