2. Four points in space $A, B, C, D$ satisfy $|AB|=3$, $|BC|=7$, $|CD|=11$, $|DA|=9$. Then the value of $AC \cdot BD$ is ( ).
(A) only one
(B) two
(C) four
(D) infinitely many
2.A.

Notice that $3^{2}+11^{2}=130=7^{2}+9^{2}$. Since $A B+B C+$ $C D+D A=0$, then
$$
\begin{array}{l}
D A^{2}=D A^{2}=(A \boldsymbol{B}+\boldsymbol{B C}+\boldsymbol{C D})^{2} \\
=A B^{2}-B C^{2}+C D^{2}+2(\boldsymbol{A B}+\boldsymbol{B C}) \cdot(B C+C D), \\
\text { i.e., } 2 A C \cdot B D=A D^{2}+B C^{2}-A B^{2}-C D^{2}=0 .
\end{array}
$$

Therefore, $A C \cdot B D$ has only one value 0.