2. Given three planes $\alpha // \beta // \gamma, \beta$ is between $\alpha$ and $\gamma, \alpha$ and $\beta$ are a distance of $4a$ apart, $\beta$ and $\gamma$ are a distance of $2a$ apart, and there is an equilateral triangle $ABC$ with side length $9a$. $A$ is in plane $\alpha$, $B$ is in plane $\beta$, $C$ is in plane $\gamma$, and $AC$ intersects plane $\beta$ at $D$. Then the acute angle formed by the plane containing $\triangle ADC$ and the three planes is $\qquad$
2. $\arcsin \frac{4 \sqrt{21}}{27}$.

As shown in the figure, draw $A O \perp \beta$ at $O$, and $A O^{\prime} \perp \gamma$ at $O^{\prime}$, then $A O=4 a, A O^{\prime}=6 a$. Draw $O E \perp B D$ at $E$, and connect $A E$, then $\angle A E O$ is the plane angle of the dihedral angle to be found.
From $\frac{A D}{A C}=\frac{A O}{A O^{\prime}}=\frac{4 a}{6 a}$ and $A C=9 a$, we get $A D=6 a$. By the cosine rule,
$$
\begin{aligned}
B D & =\sqrt{(9 a)^{2}+(6 a)^{2}-2 \cdot 9 a \cdot 6 a \cos 60^{\circ}} \\
& =3 \sqrt{7} a .
\end{aligned}
$$

From $A E \cdot B D=A B \cdot A D \cdot \sin 60^{\circ}$, we get
$$
A E=\frac{9 \sqrt{21}}{7} .
$$

Thus, $\sin A E O=\frac{A O}{A E}=\frac{4 \sqrt{21}}{27}$,
$$
\angle A E O=\arcsin \frac{4 \sqrt{21}}{27} .
$$