7. Given a tetrahedron $ABCD$ such that $AB \perp BC, BC \perp CD, AB=BC=CD=2\sqrt{3}$, and the volume of the tetrahedron is 6, then the angle between the skew lines $AD$ and $BC$ is $\qquad$.
Given that $S_{\triangle B C D}=\frac{1}{2} B C^{2}=6$, then the height of the tetrahedron $A B C D$ is 3. Establish a coordinate system $B-x y z$ as shown in the figure.
Thus, $A_{1}(0,-\sqrt{3}, 3), A_{2}(0, \sqrt{3}, 3), C(2 \sqrt{3}, 0,0)$,
$D(2 \sqrt{3}, 2 \sqrt{3}, 0)$, hence $\overrightarrow{B C}=(2 \sqrt{3}, 0,0)$,
$$
\overrightarrow{A_{1} D}=(2 \sqrt{3}, 3 \sqrt{3},-3), \overrightarrow{A_{2} D}=(2 \sqrt{3}, \sqrt{3},-3) \text {. }
$$

Then $\cos \left\langle\overrightarrow{A_{1} D}, \overrightarrow{B C}\right\rangle=\frac{12}{4 \sqrt{3} \cdot 2 \sqrt{3}}=\frac{1}{2}, \cos \left\langle\overrightarrow{A_{2} D}, \overrightarrow{B C}\right\rangle=\frac{12}{2 \sqrt{6} \cdot 2 \sqrt{3}}=\frac{\sqrt{2}}{2}$,
so the angle between the skew lines $A D$ and $B C$ is $60^{\circ}$ or $45^{\circ}$.