Convex regular 3D polytopes in R^3 composed of only regular pentagons and hexagons.

Exactly 12 pentagons: Euler's formula in $\rm I\!R^3$ is $\text{Vertices}-\text{Edges}+\text{Faces}=2$ . Two faces meet at an edge. Three faces meet at a vertex because the polygons' vertex angles prohibit any other number. Therefore, $\frac{6 H+5 P}{3}-\frac{6 H+5 P}{2}+(P+H)=2$ simplifies to $P=12$ .

Not all hexagons: Hexagons tiles the plane. The shape would never close. Therefore, it would not be convex.

Not just one hexagon: The existence proof for no hexagons is the dodecahedron. This argument comes from the dihedral angles between a pentagon and a hexagon, if one vertex of the edge is completed with specifically a type of polygon, then the other vertex has to have the same type of polygon to have edge closure of the plane polygons. To have only one hexagon, that hexagon would have to be surrounded by pentagons. The only way to complete closure of all the edges would be to have a hexagon on the other side of the polyhedron and that would mean two hexagons.

Therefore, all of the answers is the correct answer.