3. In circle $\odot O$, the radius $r=5 \mathrm{~cm}$, $A B$ and $C D$ are two parallel chords, and $A B=8 \mathrm{~cm}, C D=6 \mathrm{~cm}$. Then the length of $A C$ has ( ).
(A) 1 solution
(B) 2 solutions
(C) 3 solutions
(D) 4 solutions
3. (c).

According to the theorem and the Pythagorean theorem, the distance between $AB$ and $CD$ can be found to be $7 \mathrm{~cm}$ or $1 \mathrm{~cm}$. As shown in Figure 2, we solve for four cases respectively:
$$
\begin{array}{c}
A C_{1}=\sqrt{7^{2}+1^{2}} \\
=5 \sqrt{2}, \\
A C_{2}=\sqrt{7^{2}+7^{2}}=7 \sqrt{2}, \\
A C_{3}=\sqrt{1^{2}+1^{2}}=\sqrt{2}, \\
A C_{4}=\sqrt{1^{2}+7^{2}}=5 \sqrt{2} .
\end{array}
$$

Thus, the length of $AC$ has 3 solutions: $\sqrt{2}, 5 \sqrt{2}, 7 \sqrt{2}$.