3. As shown in the figure, in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, $A B=a, B C=b, A_{1} A=c$, and $E$ is the midpoint of $D_{1} C_{1}$. If the dihedral angle formed by plane $A_{1} B C_{1}$ and plane $A C E$ is $\theta$, then $\sin \theta=$ $\qquad$ .
3. $\frac{a^{2} b c}{\sqrt{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}} \cdot \sqrt{a^{2} b^{2}+4 b^{2} c^{2}+4 c^{2} a^{2}}}$ Detailed explanation: Let the height from $D_{1}$ to $AC$ (i.e., the distance from $D_{1}$ to $AC$) in $\triangle D_{1} A C$ be $h$, then $h=\frac{2 S_{\triangle D_{1} A C}}{a}=\frac{\sqrt{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}}}{a}$. Also, we find that $S_{\triangle E A C}=\frac{1}{2} \sqrt{a^{2} b^{2}+4 b^{2} c^{2}+4 c^{2} a^{2}}$. Let the distance from $C$ to the plane $D_{1} A E$ be $d$, then, by $V_{D_{1}-E A C}=V_{A-E D_{1} c}$, we get
$$
\begin{array}{l}
\frac{1}{3} \cdot \frac{1}{4} \sqrt{a^{2} b^{2}+4 b^{2} c^{2}+4 c^{2} a^{2}} d=\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} a c \cdot b, \quad \therefore d=\frac{a b c}{\sqrt{a^{2} b^{2}+4 b^{2} c^{2}+4 c^{2} a^{2}}} . \\
\therefore \sin \theta=\frac{d}{h}=\frac{a^{2} b c}{\sqrt{b^{2} c^{2}+c^{2} a^{2}+a^{2} b^{2}} \cdot \sqrt{a^{2} b^{2}+4 b^{2} c^{2}+4 c^{2} a^{2}}} .
\end{array}
$$