Prove that there does not exist a convex polyhedron with three triangles, six pentagons, and five heptagons. 
Prove that there does not exist a convex polyhedron with three triangles, six pentagons, and five heptagons.

Suppose on the contrary that such a convex polyhedron exists. We know that every convex polyhedron can be represented as a planar graph. Given a planar graph of such, with V the number of vertices, E the number of edges, and F the number of faces, we have that $2E$ is the sum of the number of vertices of each face. So such a planar graph has $2E = 3 \times 3 + 6 \times 5 + 5 \times 7 = 74$ so that $E = 37$. Since $F = 3 + 6 + 5 = 14$, by Euler's formula we have $V = 2 + E - F = 25$. Now, by doubling each edge (i.e. replace each edge by 2 edges connecting the same end vertices) and counting, we know that $2 \sum \deg v = 2E$ = 74, so that $\sum \deg v = 37$. As there are $25$ vertices and each vertex has $\deg v \geq 1$, it must be that some vertices must have $\deg v = 1$, which is impossible since our planar graph comes from a convex polyhedron.
Is this proof correct?

I don't follow your "double each edge and count" step.  It seems to me that you should have
$$\sum_{v} \deg v = 74 < 3V$$
implying that some vertex has degree at most $2$, which cannot happen in a convex polyhedron.
Otherwise, it looks fine to me.
