$1 \cdot 44$ Given $a=1990 x+1989, b=1990 x+1990, c=$ $1990 x+1991$. Then, the value of $a^{2}+b^{2}+c^{2}-a b-b c-c a$ is
(A) 0 .
(B) 1 .
(C) 2 .
(D) 3 .
(China Guangzhou, Wuhan, Fuzhou, and other five cities junior high school mathematics league, 1990)
[Solution 1] From the given, we know $a=b-1, c=b+1$. Then
$$
\begin{aligned}
\text { Original expression }= & (b-1)^{2}+b^{2}+(b+1)^{2}-(b-1) b-b(b+1)-(b-1)(b \\
& +1) \\
= & 3 .
\end{aligned}
$$

Therefore, the answer is $(D)$.
[Solution 2] Original expression $=\frac{1}{2}\left[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}\right]$
$$
=\frac{1}{2}(1+1+4)=3 \text {. }
$$

Therefore, the answer is $(D)$.