6. In the tetrahedron $ABCD$, $AB \perp CD, AB=a, AB$ forms an angle $\alpha$ with the plane $BCD$, and the dihedral angle $A-CD-B$ is $\beta$. Then the minimum value of the perimeter of the triangular cross-section of the tetrahedron passing through the edge $AB$ is
6. $-\frac{a}{\sin \beta}[\sin \alpha+\sin \beta+\sin (\alpha+\beta)]$

As shown in the figure, draw $A E \perp C D$ at $E$, and connect $B E$. Since $A B \perp C D$, we have $C D \perp$ plane $A B E$, so $C D \perp B E$, and $\angle A B E$ is the angle between $A B$ and plane $B C D$. Therefore, $\angle A B E=\alpha$; and $\angle A E B$ is the plane angle of the dihedral angle $A-C D-B$, so $\angle A E B=\beta$.
It is easy to see that the perimeter of $\triangle A B E$ is the smallest.
By the Law of Sines, we have $\frac{A B}{\sin \beta}=\frac{A E}{\sin \alpha}=\frac{B E}{\sin (\alpha+\beta)}$.
By the properties of proportions, we have $\frac{A B+A E+B E}{\sin \beta+\sin \alpha+\sin (\alpha+\beta)}=\frac{A B}{\sin \beta}=\frac{a}{\sin \beta}$.
Therefore, $A B+A E+B E=\frac{a}{\sin \beta}[\sin \alpha+\sin \beta+\sin (\alpha+\beta)]$.