Answer: 321

Solution:

Since $2 - 2x + x^2 = (x - 1)^2 + 1$ $= (x - (1+i))(x - (1-i)),$ each factor of the product splits, and as $k$ runs from $0$ to $12,$ $\omega^k$ runs over all $13$th roots of unity. Because $\prod_k (x - \omega^k) = x^{13} - 1,$ for any $\alpha$ we get $\prod_k (\omega^k - \alpha) = (-1)^{13}(\alpha^{13} - 1)$ $= 1 - \alpha^{13}.$ Hence the product equals $$\left(1 - (1+i)^{13}\right)\left(1 - (1-i)^{13}\right).$$

Since $(1+i)^2 = 2i,$ we get $(1+i)^{13} = (1+i)(2i)^6$ $= -64(1 + i) = -64 - 64i,$ and by conjugation $(1-i)^{13} = -64 + 64i.$ So the product is $$\begin{gathered} (65 + 64i)(65 - 64i) \\ = 65^2 + 64^2 = 4225 + 4096 \\ = 8321, \end{gathered}$$ whose remainder upon division by $1000$ is $321.$