Having a sphere and three diffrent point $A,B,C$ on this sphere. Find set of all centers of spheres inscribed in a tetrahedron $ABCD$, where $D$ is some point on the given sphere. The problem reduced to 2dimensions is trivial it's just sum of two arcs of some circle, but in 3dimensions the set is not so simple. Checking in geogebra it's not sum of some parts of sphere. I don't know what this set looks and how it's described.

1$\begingroup$ Why restrict $A$, $B$, and $C$ to be on the sphere? $\endgroup$ – Douglas Zare Aug 15 '15 at 16:08

1$\begingroup$ The boundary of the set is formed by incenters of tetrahedra with $D$ at infinity. Do not expect it to be particular nice surface. $\endgroup$ – Anton Petrunin Aug 15 '15 at 16:33

1$\begingroup$ $D$ is on the same sphere as $A,B,C$. It makes that the set is inside that sphere $\endgroup$ – M.Martin Aug 15 '15 at 19:29

$\begingroup$ @AntonPetrunin: "with $D$ at infinity": Could you expand on your comment? $\endgroup$ – Joseph O'Rourke Aug 15 '15 at 21:39

$\begingroup$ @JosephO'Rourke, ignore it, it was a comment to the old version of the question where $D$ is arbitrary. $\endgroup$ – Anton Petrunin Aug 16 '15 at 14:15
A little algebra shows that, for $A=(1,0,0)$, $B=(0,1,0)$, and $C=(0,0,1)$ on the sphere, this surface is an irreducible algebraic surface of degree 5 that is singular at $A$, $B$, and $C$ but is otherwise smooth (even on the plane at infinity). In fact, the hypersurface is defined by $$ \begin{aligned} 0 &= 2s_1^59s_1^3s_2+9s_1^2s_3+9s_1s_2^227s_2s_3\\ &\quad 7s_1^4+18s_1^2s_2+18s_1s_39s_2^2\\ &\quad\quad +8s_1^39s_1s_227s_3 2s_1^2 2s_1 +1 \end{aligned} $$ where $s_1=x+y+z$, $s_2=xy+yz+zx$, and $s_3=xyz$. (Written out in terms of $x$, $y$, and $z$, the polynomial on the right hand side has 56 terms; as Anton suspected, it does not appear to be very nice.)
The singularities at $A$, $B$, and $C$ are cubic (i.e., the polynomial, when expanded about each of these points, has lowest nonvanishing terms of order $3$), so one suspects that this surface, once the singularities have been resolved, might be a known algebraic surface. In fact, the projectivization of the tangent cone at each of these three points is a nonsingular cubic curve. (I have not computed the jinvariant, though.)
For $A,B,C=$ $(1,0,0)$, $(0,1,0)$, $(0,0,1)$ (blue below) on a unit sphere $S$, the surface is a sort of triangular tea bag with corners at $A,B,C$. Below are two views of $100$ random tetrahedron incenters (red) on the surface, corresponding to random points $D$ (not shown) uniformly distributed on $S$.
A typical $D$ (green), the determined tetrahedron, the inscribed sphere and its (red) center, are shown below.
Here is a depiction of Robert Bryant's surface defined by the $56$term polynomial he details:
Note one component is the "triangular tea bag" discernable in my empirical investigation. Here is the polynomial: $$ 2 x^5+x^4 y+x^4 z7 x^4+2 x^3 y^2+4 x^3 y z10 x^3 y+2 x^3 z^210 x^3 z+8 x^3+2 x^2 y^312 x^2 y^2 z15 x^2 y^212 x^2 y z^2+6 x^2 y z+15 x^2 y+2 x^2 z^315 x^2 z^2+15 x^2 z2 x^2+x y^4+4 x y^3 z10 x y^312 x y^2 z^2+6 x y^2 z+15 x y^2+4 x y z^3+6 x y z^26 x y z4 x y+x z^410 x z^3+15 x z^24 x z2 x+2 y^5+y^4 z7 y^4+2 y^3 z^210 y^3 z+8 y^3+2 y^2 z^315 y^2 z^2+15 y^2 z2 y^2+y z^410 y z^3+15 y z^24 y z2 y+2 z^57 z^4+8 z^32 z^22 z+1$$ The plot above restricts $(x,y,z)$ to lie on or in the sphere.

2$\begingroup$ Thanks for making the drawing. In case others might wonder what the other 'components' are, here is a comment: What the algebraic surface describes is the locus of centers of spheres that are tangent to all four planes that make up the sides of a tetrahedron ABCD where D is an arbitrary point on the 2sphere. Of course, one such sphere will be the sphere that lies in the convex hull of the 4 points, but, of course, there are four others. The polynomial equation can't distinguish these. As your picture shows, even requiring that the center be interior to the original sphere is not enough. $\endgroup$ – Robert Bryant Aug 17 '15 at 3:31