Let $ ABC$ be a triangle with $ AB \equal{} AC$. A ray $ Ax$ is constructed in space such that the three planar angles of the trihedral angle $ ABCx$ at its vertex $ A$ are equal. If a point $ S$ moves on $ Ax$, find the locus of the incenter of triangle $ SBC$.
1. **Given Conditions and Setup:**
   - We have a triangle \( ABC \) with \( AB = AC \), making it an isosceles triangle.
   - A ray \( Ax \) is constructed such that the three planar angles of the trihedral angle \( ABCx \) at its vertex \( A \) are equal.
   - A point \( S \) moves on \( Ax \).
   - We need to find the locus of the incenter of triangle \( SBC \).

2. **Understanding the Trihedral Angle:**
   - Since the three planar angles at \( A \) are equal, each angle is \( 120^\circ \). This implies that the ray \( Ax \) is such that the angles \( \angle BAC = \angle BAX = \angle CAX = 120^\circ \).

3. **Incenter of Triangle \( SBC \):**
   - Let \( I \) be the incenter of triangle \( ABC \).
   - Let \( I' \) be the incenter of triangle \( S'BC \) where \( S' \) is a point at infinity on \( Ax \).

4. **Locus of the Incenter:**
   - The locus of the incenter of triangle \( SBC \) is part of a circle with center \( O \) and radius \( r \).
   - This circle lies on the plane perpendicular to the plane of triangle \( ABC \) and passes through \( A \) and \( M \), where \( M \) is the midpoint of \( BC \).

5. **Finding the Center \( O \):**
   - The center \( O \) is the intersection of the perpendicular bisector of \( II' \) (where \( I \) is the incenter of triangle \( ABC \) and \( I' \) is the incenter of triangle \( S'BC \)) and the line \( My \) (where \( \angle AMy = 90^\circ + \beta \) with \( \beta = \angle SAM \)).

6. **Radius of the Circle:**
   - The radius \( r \) of the circle is given by:
     \[
     r = \frac{d}{c \sin \beta} \sqrt{b^2 + c^2 - 2bc \cos \beta}
     \]
     where:
     - \( 2b = BC \)
     - \( c = IM \)
     - \( 2d = II' \)
     - \( \alpha = \angle BAC \)
     - \( \beta = \sin^{-1} \left( \sec \frac{\alpha}{2} \sqrt{\sin^2 \alpha - \sin^2 \frac{\alpha}{2}} \right) \)

7. **Applying Inversion:**
   - To solve the problem using inversion, we consider an inversion centered at \( A \) with an appropriate radius. This transformation will map the line \( Ax \) to a circle passing through \( A \) and the points \( B \) and \( C \) to other points on this circle.
   - The incenter of the inverted triangle will trace a circle as \( S \) moves along \( Ax \).

8. **Conclusion:**
   - The locus of the incenter of triangle \( SBC \) is part of a circle from \( I \) to \( I' \) with the given radius \( r \).

The final answer is \(\boxed{\text{part of a circle with center } O \text{ and radius } r}\).