Insphere::usage="Given the vertices of a tetrahedron, the incenter and the inradius of the inesphere are obtained.​​You just have to provide the vertices of a tetrahedron in the command below, evaluate notebook.​Reference: Personal.";
In[]:=
​​​​NotableSphere[tet={{ax_,ay_,az_},{bx_,by_,bz_},{cx_,cy_,cz_},{dx_,dy_,dz_}}]:=​​Module[{},prueba=Det[{{1,ax,ay,az},{1,bx,by,bz},{1,cx,cy,cz},{1,dx,dy,dz}}];​​If[prueba==0,Print[PasteButton["It's not a tetrahedron"]],​​ρ=Insphere[tet];​​s=Grid[{{Incenter,Inradius},{ρ[[1]],ρ[[2]]}},FrameAll];​​Print[s];​​Show[{Graphics3D[{Tube[#]&/@Subsets[tet,{2}],Black,Sphere[ρ[[1]],0.03],Opacity[.1],Green,Polygon/@Subsets[tet,{3}]}],Graphics3D[{Opacity[0.6],Sphere[ρ[[1]],ρ[[2]]]}]},AxesLabel->(Style[#,15,Blue]&/@{"X","Y","Z"}),AxesOrigin{0,0,0},AxesTrue,BoxedFalse,BoxRatiosAutomatic,PlotRangeAll]​​]]
In[]:=
NotableSpheretet=
2
2
,0,-
6
6
,-
2
2
,0,-
6
6
,0,
2
,-
6
6
,0,0,
6
3

Incenter
Inradius
0,
3
2
3+2
3
+
57
,
4
2
-
6
-
38
2(3+2
3
+
57
)

3
2
3+2
3
+
57
Out[]=