1. Let $a, b, c$ be three distinct positive integers. If
$$
\begin{array}{l}
\{a+b, b+c, c+a\} \\
=\left\{n^{2},(n+1)^{2},(n+2)^{2}\right\}\left(n \in \mathbf{N}_{+}\right),
\end{array}
$$
then the minimum value of $a^{2}+b^{2}+c^{2}$ is ( ).
(A) 2007
(B) 1949
(C) 1297
(D) 1000
- 1.C.

Assume $a>b>c$, then $a+b>c+a>b+c$.
Since $(a+b)+(b+c)+(c+a)=2(a+b+c)$ is an even number, $n^{2} 、(n+1)^{2} 、(n+2)^{2}$ must be two odd and one even. Therefore, $n$ must be an odd number.
Also, since $b+c>1$, $n$ must be an odd number no less than 3.
If $n=3$, then $\{a+b, b+c, c+a\}=\left\{3^{2}, 4^{2}, 5^{2}\right\}$.
Thus, $a+b+c=\frac{1}{2}\left(3^{2}+4^{2}+5^{2}\right)=5^{2}$, and $a+b=5^{2}$.
Therefore, $c=0$, which does not meet the requirement.
If $n=5$, then $\{a+b, b+c, c+a\}=\left\{5^{2}, 6^{2}, 7^{2}\right\}$.
Thus, $\left\{\begin{array}{l}a+b=7^{2}, \\ c+a=6^{2} \\ b+c=5^{2} .\end{array}\right.$ Solving this, we get $\left\{\begin{array}{l}a=30, \\ b=19, \\ c=6 .\end{array}\right.$
At this point, $a^{2}+b^{2}+c^{2}=1297$.