(A.Zaslavsky) Given three points $ C_0$, $ C_1$, $ C_2$ on the line $ l$. Find the locus of incenters of triangles $ ABC$ such that points $ A$, $ B$ lie on $ l$ and the feet of the median, the bisector and the altitude from $ C$ coincide with $ C_0$, $ C_1$, $ C_2$.
1. **Restate the problem with new notation:**
   Given three points \( M, V, D \) on the line \( \ell \). We need to find the locus of incenters of triangles \( ABC \) such that points \( A \) and \( B \) lie on \( \ell \) and the feet of the median, the bisector, and the altitude from \( C \) coincide with \( M, V, D \) respectively.

2. **Define the incenters and excenters:**
   Let \( I \) be the incenter and \( I_c \) be the \( C \)-excenter of \( \triangle ABC \). Let \( X \) and \( Y \) be the tangency points of the incircle \((I)\) and the \( C \)-excircle \((I_c)\) with the line \( \ell \equiv AB \).

3. **Harmonic separation:**
   Since the points \( C, V, I, I_c \) are harmonically separated, their orthogonal projections \( D, V, X, Y \) onto \( \ell \) are also harmonically separated. This means that the cross-ratio of these four points is \(-1\).

4. **Midpoint property:**
   The point \( M \) is the midpoint of segment \( XY \). By Newton's theorem, we have:
   \[
   MX^2 = MY^2 = MV \cdot MD = \text{const}
   \]
   This implies that the distance from \( M \) to \( X \) (and \( Y \)) is constant.

5. **Fixed point \( X \):**
   Since \( MX^2 = \text{const} \), the point \( X \) is fixed between \( D \) and \( V \).

6. **Locus of \( I \):**
   Therefore, the locus of the incenter \( I \) is the perpendicular line \( \tau \) to \( \ell \) through the fixed point \( X \).

\(\blacksquare\)

The final answer is the perpendicular line \( \boxed{ \tau } \) to \( \ell \) through the fixed point \( X \).