$[\quad$ Volume of a parallelepiped $\quad]$

In the base of a rectangular parallelepiped lies a square with side $2 \sqrt{3}$. The diagonal of one lateral face forms an angle of $30^{\circ}$ with the plane of the adjacent lateral face. Find the volume of the parallelepiped.
Let the diagonal $A B 1$ of the lateral face $A A 1 B 1 B$ of the rectangular parallelepiped $A B C D A 1 B 1 C 1 D 1$ form an angle of $30^{\circ}$ with the plane of the lateral face $B B 1 C 1 C$. The line $A B$ is perpendicular to the plane of the face $B B 1 C 1 C$, since it is perpendicular to two intersecting lines $B B 1$ and $B C$ of this plane. Therefore, $B B 1$ is the orthogonal projection of the inclined line $A B 1$ onto the plane of the face $B B 1 C 1 C$. Thus, $A B 1 B$ is the angle between the line $A B 1$ and the plane of the face $B B 1 C 1 C$. According to the problem, $\angle A B 1 B=30^{\circ}$. From the right triangle $A B 1 B$, we find that

$$
B B 1=A B \operatorname{ctg} \angle A B 1 B=2 \sqrt{3} \cdot \sqrt{3}=6 \text {. }
$$

Therefore,

$$
V_{A B C D A} 1 B 1 C 1 D 1=S_{A B C D} \cdot B B 1=(2 \sqrt{3}) 2 \cdot 6=72
$$

## Answer

72.00