Consider a trihedral angle $Sxyz$ with $\angle xSz = \alpha, \angle xSy = \beta$ and $\angle ySz =\gamma$. Let $A,B,C$ denote the dihedral angles at edges $y, z, x$ respectively.
$(a)$ Prove that $\frac{\sin\alpha}{\sin A}=\frac{\sin\beta}{\sin B}=\frac{\sin\gamma}{\sin C}$
$(b)$ Show that $\alpha + \beta = 180^{\circ}$ if and only if $\angle A + \angle B = 180^{\circ}.$
$(c)$ Assume that $\alpha=\beta =\gamma = 90^{\circ}$. Let $O \in Sz$ be a fixed point such that $SO = a$ and let $M,N$ be variable points on $x, y$ respectively. Prove that $\angle SOM +\angle SON +\angle MON$ is constant and find the locus of the incenter of $OSMN$.
### Part (a)
We need to prove that:
\[
\frac{\sin \alpha}{\sin A} = \frac{\sin \beta}{\sin B} = \frac{\sin \gamma}{\sin C}
\]

1. Consider a trihedral angle \( Sxyz \) with \(\angle xSz = \alpha\), \(\angle xSy = \beta\), and \(\angle ySz = \gamma\). Let \( A, B, C \) denote the dihedral angles at edges \( y, z, x \) respectively.

2. Take a point \( M \) on \( Sx \), project \( M \) onto the plane \( Syz \) and let \( N \) be the projection of \( M \). Let \( P \) and \( Q \) be the projections of \( M \) onto \( Sy \) and \( Sz \) respectively.

3. By the three perpendicular lines theorem, \( NP \) is perpendicular to \( Sy \) and \( NQ \) is perpendicular to \( Sz \).

4. We have:
   \[
   MN = MP \sin \angle ySz = MS \sin \gamma \sin A
   \]
   \[
   MN = MQ \sin \angle xSy = MS \sin \beta \sin C
   \]

5. Thus, we get:
   \[
   \frac{\sin \beta}{\sin B} = \frac{\sin \gamma}{\sin C}
   \]

6. Similarly, we can show:
   \[
   \frac{\sin \alpha}{\sin A} = \frac{\sin \gamma}{\sin C}
   \]

7. Therefore, we conclude:
   \[
   \frac{\sin \alpha}{\sin A} = \frac{\sin \beta}{\sin B} = \frac{\sin \gamma}{\sin C}
   \]

### Part (b)
We need to show that \(\alpha + \beta = 180^\circ\) if and only if \(\angle A + \angle B = 180^\circ\).

1. Assume \(\alpha + \beta = 180^\circ\). Then \(\gamma = 0^\circ\).

2. From part (a), we have:
   \[
   \frac{\sin \alpha}{\sin A} = \frac{\sin \beta}{\sin B}
   \]

3. Since \(\alpha + \beta = 180^\circ\), \(\sin \alpha = \sin (180^\circ - \beta) = \sin \beta\).

4. Therefore, \(\sin A = \sin B\), which implies \(A = 180^\circ - B\).

5. Conversely, assume \(\angle A + \angle B = 180^\circ\). Then \(\sin A = \sin B\).

6. From part (a), we have:
   \[
   \frac{\sin \alpha}{\sin A} = \frac{\sin \beta}{\sin B}
   \]

7. Since \(\sin A = \sin B\), \(\sin \alpha = \sin \beta\), which implies \(\alpha = 180^\circ - \beta\).

8. Therefore, \(\alpha + \beta = 180^\circ\).

### Part (c)
Assume \(\alpha = \beta = \gamma = 90^\circ\). Let \(O \in Sz\) be a fixed point such that \(SO = a\) and let \(M, N\) be variable points on \(x, y\) respectively. We need to prove that \(\angle SOM + \angle SON + \angle MON\) is constant and find the locus of the incenter of \(OSMN\).

1. Since \(\alpha = \beta = \gamma = 90^\circ\), the trihedral angle is a right trihedral angle.

2. The angles \(\angle SOM\), \(\angle SON\), and \(\angle MON\) are all right angles.

3. Therefore, \(\angle SOM + \angle SON + \angle MON = 90^\circ + 90^\circ + 90^\circ = 270^\circ\).

4. The incenter of \(OSMN\) is the point equidistant from all sides of the triangle. Since \(O\) is fixed and \(M, N\) are variable points on \(x, y\) respectively, the incenter will lie on the angle bisector of \(\angle SOM\) and \(\angle SON\).

5. The locus of the incenter is a circle centered at \(O\) with radius \(a\).

The final answer is \(\boxed{\frac{\sin \alpha}{\sin A} = \frac{\sin \beta}{\sin B} = \frac{\sin \gamma}{\sin C}}\)