2. Given $a=1999 x+2000, b=1999 x+2001, c$ $=1999 x+2002$. Then the value of the polynomial $a^{2}+b^{2}+c^{2}-a b-b c-$ $c a$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
2. (D).
$$
\begin{array}{l}
\because a^{2}+b^{2}+c^{2}-a b-b c-c a \\
=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right] . \\
\text { Also } a-b=-1, b-c=-1, c-a=2, \\
\therefore \text { the original expression }=\frac{1}{2}\left[(-1)^{2}+(-1)^{2}+2^{2}\right]=3 .
\end{array}
$$