7. Given a square $A B C D$ with side length $a$, vertex $A$ lies in plane $\beta$, and the other vertices are on the same side of plane $\beta$. The distances from points $B$ and $D$ to plane $\beta$ are $1$ and $2$, respectively. If the angle between plane $A B C D$ and plane $\beta$ is $30^{\circ}$, then $a=$
$7.2 \sqrt{5}$. 
As shown in Figure 3, draw perpendiculars from points $B$ and $D$ to plane $\beta$, intersecting plane $\beta$ at points $B_{1}$ and $D_{1}$, respectively.
Then $B B_{1}=1$,
$D D_{1}=2$.
Let $A D=a$. Then
$$
\begin{array}{l}
A B_{1}=\sqrt{a^{2}-1}, A D_{1}=\sqrt{a^{2}-4}, \\
B_{1} D_{1}=\sqrt{2 a^{2}-1} .
\end{array}
$$

Thus, $\cos \angle D_{1} A B_{1}=\frac{2}{\sqrt{a^{2}-1} \sqrt{a^{2}-4}}$
$$
\begin{array}{l}
\Rightarrow \sin \angle D_{1} A B_{1}=\frac{\sqrt{a^{4}-5 a^{2}}}{\sqrt{a^{2}-1} \sqrt{a^{2}-4}} \\
\Rightarrow S_{\triangle D_{1} A B_{1}}=\frac{1}{2} \sqrt{a^{4}-5 a^{2}} \\
\Rightarrow \frac{\sqrt{a^{4}-5 a^{2}}}{a^{2}}=\frac{\sqrt{3}}{2} \Rightarrow a=2 \sqrt{5} .
\end{array}
$$