Points $ A’,B,C’ $ are arbitrarily taken on edges $ SA,SB, $ respectively, $ SC $ of a tetrahedron $ SABC. $ Plane forrmed by $ ABC $ intersects the plane $ \rho , $ formed by $ A’B’C’, $ in a line $ d. $ Prove that, meanwhile the plane $ \rho $ rotates around $ d, $ the lines $ AA’,BB’ $ and $ CC’ $ are, and remain concurrent. Find de locus of the respective intersections.
1. **Define the points and planes:**
   - Let \( A', B', C' \) be points on the edges \( SA, SB, SC \) of the tetrahedron \( SABC \), respectively.
   - The plane formed by \( A, B, C \) is denoted as \( \pi \).
   - The plane formed by \( A', B', C' \) is denoted as \( \rho \).
   - The intersection of planes \( \pi \) and \( \rho \) is a line \( d \).

2. **Concurrent lines:**
   - We need to prove that the lines \( AA', BB', CC' \) are concurrent and remain concurrent as the plane \( \rho \) rotates around the line \( d \).

3. **Intersection of lines:**
   - Consider the lines \( AA', BB', CC' \). Since \( A', B', C' \) are on the edges \( SA, SB, SC \), these lines intersect at point \( S \).

4. **Rotation of plane \( \rho \):**
   - When the plane \( \rho \) rotates around the line \( d \), the points \( A', B', C' \) move along the edges \( SA, SB, SC \), respectively.
   - However, the lines \( AA', BB', CC' \) will still intersect at point \( S \) because \( A', B', C' \) remain on the edges \( SA, SB, SC \).

5. **Locus of intersection:**
   - The locus of the intersection of the lines \( AA', BB', CC' \) is the point \( S \), as the lines always intersect at \( S \) regardless of the rotation of the plane \( \rho \).

\[
\boxed{S}
\]