Three mutually perpendicular rays $O_x,O_y,O_z$ and three points $A,B,C$ on $O_x,O_y,O_z$, respectively. A variable sphere є through $A, B,C$ meets $O_x,O_y,O_z$ again at $A', B',C'$, respectively. Let $M$ and $M'$ be the centroids of triangles $ABC$ and $A'B'C'$. Find the locus of the midpoint of $MM'$.
1. Let the coordinates of points \( A, B, C \) be \( (a, 0, 0), (0, b, 0), (0, 0, c) \) respectively. These points lie on the coordinate axes \( O_x, O_y, O_z \).
2. The sphere passing through \( A, B, C \) intersects the coordinate axes again at points \( A', B', C' \) with coordinates \( (a', 0, 0), (0, b', 0), (0, 0, c') \) respectively.
3. The centroid \( M \) of triangle \( ABC \) is given by:
   \[
   M = \left( \frac{a}{3}, \frac{b}{3}, \frac{c}{3} \right)
   \]
4. The centroid \( M' \) of triangle \( A'B'C' \) is given by:
   \[
   M' = \left( \frac{a'}{3}, \frac{b'}{3}, \frac{c'}{3} \right)
   \]
5. The midpoint \( S \) of segment \( MM' \) is given by:
   \[
   S = \left( \frac{\frac{a}{3} + \frac{a'}{3}}{2}, \frac{\frac{b}{3} + \frac{b'}{3}}{2}, \frac{\frac{c}{3} + \frac{c'}{3}}{2} \right)
   \]
   Simplifying, we get:
   \[
   S = \left( \frac{a + a'}{6}, \frac{b + b'}{6}, \frac{c + c'}{6} \right)
   \]
6. Let \( A'', B'', C'' \) be the midpoints of \( AA', BB', CC' \) respectively. Then:
   \[
   A'' = \left( \frac{a + a'}{2}, 0, 0 \right), \quad B'' = \left( 0, \frac{b + b'}{2}, 0 \right), \quad C'' = \left( 0, 0, \frac{c + c'}{2} \right)
   \]
7. The coordinates of \( S \) can be rewritten as:
   \[
   S = \left( \frac{a''}{3}, \frac{b''}{3}, \frac{c''}{3} \right)
   \]
8. Since \( A'', B'', C'' \) are the midpoints of \( AA', BB', CC' \), they lie on the coordinate axes. Therefore, the coordinates of \( S \) are:
   \[
   S = \left( \frac{x}{3}, \frac{y}{3}, \frac{z}{3} \right)
   \]
   where \( x, y, z \) are the coordinates of the midpoints of \( AA', BB', CC' \) respectively.

Conclusion:
The locus of the midpoint of \( MM' \) is a line passing through the origin and is given by:
\[
S = \left( \frac{x}{3}, \frac{y}{3}, \frac{z}{3} \right)
\]

The final answer is \( \boxed{ \left( \frac{x}{3}, \frac{y}{3}, \frac{z}{3} \right) } \).