12.1 .3 * Four points in space $A, B, C, D$ satisfy $|\overrightarrow{A B}|=3,|\overrightarrow{B C}|=7,|\overrightarrow{C D}|=11$, $|\overrightarrow{D A}|=9$, then the value of $\overrightarrow{A C} \cdot \overrightarrow{B D}$ is ( ).
(A) only one
(B) two
(C) four
(D) infinitely many
Since $\overrightarrow{A B}^{2}+\overrightarrow{C D}^{2}=3^{2}+11^{2}=130=7^{2}+9^{2}=\overrightarrow{B C}^{2}+\overrightarrow{D A}^{2}$, from $\overrightarrow{A B}+$ $\overrightarrow{B C}+\overrightarrow{C D}+\overrightarrow{D A}=\overrightarrow{0}$ we get $\overrightarrow{A B}+\overrightarrow{C D}=-(\vec{C}+D \vec{A})$, squaring both sides gives $\overrightarrow{A B} \cdot \overrightarrow{C D}=\overrightarrow{B C} \cdot \overrightarrow{D A}$, hence $\overrightarrow{A B} \cdot \overrightarrow{C D}=-\overrightarrow{A D} \cdot \overrightarrow{B C}$. Therefore, $\overrightarrow{A C} \cdot \overrightarrow{B D}=(\overrightarrow{A B}+\overrightarrow{B C})(\overrightarrow{B C}+\overrightarrow{C D})=(\overrightarrow{A B}+$ $\overrightarrow{B C}+\overrightarrow{C D}) \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}=\overrightarrow{A D} \cdot \overrightarrow{B C}+\overrightarrow{A B} \cdot \overrightarrow{C D}=0$. $\overrightarrow{A C} \cdot \overrightarrow{B D}$ has only one value 0, so the answer is A.