7. (a) Suppose that $X, Y, Z$ are fixed on segments $A B, B C, C D$. It is proven in a standard way that if $\angle A T X \neq \angle Z T D$, then $Z T + T X$ can be reduced. It follows that if there exists a broken line $X Y Z T X$ of minimal length, then the following conditions hold:
$$\begin{aligned} & \angle D A B = \pi - \angle A T X - \angle A X T \\ & \angle A B C = \pi - \angle B X Y - \angle B Y X = \pi - \angle A X T - \angle C Y Z \\ & \angle B C D = \pi - \angle C Y Z - \angle C Z Y \\ & \angle C D A = \pi - \angle D T Z - \angle D Z T = \pi - \angle A T X - \angle C Z Y. \end{aligned}$$
Thus $\sigma = 0$. (b) Now let $\sigma = 0$. Let us cut the surface of the tetrahedron along the edges $A C, C D$, and $D B$ and set it down into a plane. Consider the plane figure $\mathcal{S} = A C D' B D'' C'$ thus obtained made up of triangles $B C D', A B C, A B D''$, and $A C' D''$, with $Z', T', Z''$ respectively on $C D', A D'', C' D''$ (here $C'$ corresponds to $C$, etc.). Since $\angle C' D'' A + \angle D'' A B + \angle A B C + \angle B C D' = 0$ as an oriented angle (because $\sigma = 0$), the lines $C D'$ and $C' D''$ are parallel and equally oriented; i.e., $C D' D'' C'$ is a parallelogram. The broken line $X Y Z T X$ has minimal length if and only if $Z'', T', X$, $Y, Z'$ are collinear (where $Z' Z'' \| C C'$), and then this length equals $Z' Z'' = C C' = 2 A C \sin (\alpha / 2)$. There is an infinity of such lines, one for every line $Z' Z''$ parallel to $C C'$ that meets the interiors of all the segments $C B, B A, A D''$. Such ! $Z' Z''$ exist. Indeed, the triangles $C A B$ and $D'' A B$ are acute-angled, and thus the segment $A B$ has a common interior point with the parallelogram $C D' D'' C'$. Therefore the desired result follows.