I'll restate the notation of the problem differently: Quote: (A.Zaslavsky) Given three points $ M,$ $ V,$ $ D$ on the line $ \ell.$ Find the locus of incenters of triangles $ ABC$ such that points $ A,$ $ B$ lie on $ \ell$ and the feet of the median, the bisector and the altitude from $ C$ coincide with $ M,$ $ V,$ $ D.$ Let $ I,I_c$ be the incenter and C-excenter of $ \triangle ABC.$ $X,Y$ are the tangency points of $ (I),(I_c)$ with $ \ell \equiv AB.$ Since the points $ C,V,I,I_c$ are harmonically separeted, so are their orthogonal projections $ D,V,X,Y$ onto $ \ell.$ $ M$ is also midpoint of segment $ XY,$ hence by Newton's theorem $ MX^2 = MY^2 = MV \cdot MD = \text{const}$ $ \Longrightarrow$ $ X$ is a fixed point between $ D,V.$ Therefore the locus of $ I$ is the perpendicular line $ \tau$ to $ \ell$ through $ X.$