124.2 ** Gluing the bases of two congruent regular tetrahedra together, we get a hexahedron with all dihedral angles equal, and the shortest edge of this hexahedron is 2. Then the distance between the farthest two points is $\qquad$ .
Determine the edge length of the hexahedron. As shown in the figure, draw $C E \perp A D$, connect $E F$, it is easy to prove that $E F \perp A D$, thus $\angle C E F$ is the plane angle of the dihedral angle $C-A D-F$. Let $G$ be the midpoint of $C D$, similarly, $\angle A G B$ is the plane angle of the dihedral angle $A-C D-B$. Given that $\angle C E F=\angle A G B$. Let the side length of the base $\triangle C D F$ be $2 a$, and the side length $A D$ be $b$. In $\triangle A C D$, $C E \cdot b=A G \cdot 2 a$, hence $C E=$ $\frac{A G \cdot 2 a}{b}=\frac{\sqrt{b^{2}-a^{2}} \cdot 2 a}{b}$. In $\triangle A B C$, it is easy to find that $A B=2 \sqrt{b^{2}-\frac{4}{3} a^{2}}$. From $\triangle C E F \backsim \triangle A G B$ we get $\frac{A B}{C F}=$ $\frac{A G}{C E}$, i.e., $\frac{2 \sqrt{b^{2}-\frac{4}{3} a^{2}}}{2 a}=\frac{\sqrt{b^{2}-a^{2}}}{\frac{\sqrt{b^{2}-a^{2}} \cdot 2 a}{b}}$. Solving this, we get $b=$ $\frac{4}{3} a$. Therefore, $b=2,2 a=3$ : Thus, the distance between the farthest two vertices is 3.