A triangle, two squares, five pentagons and $x$ hexagons met in the plane... Is it possible to draw a cubic planar graph consisting of:


*

*one triangle

*two squares

*five pentagons

*and an undefined amount of hexagons?


In total $F=F_3+F_4+F_5+F_6=8+F_6$. According to the formula
$$
4F_2+3F_3+2F_4+F_5+0F_6-F_7-2F_8-3F_9-\cdots = 12
$$
given here, it could be possible, since $3\cdot 1 + 2\cdot 2 + 5=12\checkmark $.
According to 
$$3F_3+4F_4+5F_5+6F_6=2E$$
the corresponding graph would have $E=18+3F_6$ edges. Since it's cubic $V=\frac23E=12+2F_6$ vertices. The last result verifies Euler's formula, since 
$$
V+F=E+2\\
12+2F_6+ 8+F_6=18+3F_6+2\checkmark
$$
It tried and searched the whole evening, without success. Is it possible that the graph doesn't exist?
 A: There seems to be only one solution. In Mathematica,
GraphData[{"Cubic", {14, 277}}]  

A: A cubic planar graph with only one triangle, two squares, five pentagons, and some hexagons cannot have the triangle surrounded by three pentagons. We can prove this by examining cases.
As we draw a graph, let's call a vertex with only two incident edges "open" (we can add another edge to it) and vertices with three edges "full".
A triangle surrounded by pentagons has an outer ring of six vertices, alternating between open and full:

If all the adjacent faces are hexagons, the outer ring of vertices remains the same:

So whether we have zero or 100 such rings of hexagons, the effect when we finally encounter a square or pentagon will be the same.
Case 1
Suppose we have a square:

Now there are two open vertices. Either there is an edge between them (Case 1a), they have entirely separate edges (Case 1b), or they each have an edge to the same vertex (Case 1c).



In case (a), there are three squares, which is too many.
In case (b), the other faces adjacent to our pentagons have more than five edges, hence must be hexagons. But this creates a digon, which is forbidden.
In case (c), we end up with a single open vertex; adding an edge to this creates a degenerate face which crosses the same edge twice, having more than six edges, which is forbidden.

Suppose we have a pentagon:

Now the 3 red edges form part of a face $F$.
Case 2
If $F$ is a square, like so:

Then all the outer edges are part of a face, which is either a pentagon with an open vertex left (impossible) or a degenerate face (crossing the same edge twice) with more than six edges (impossible).
Case 3
If $F$ is a pentagon, like so:

then the four blue edges are part of a face with more than four sides.
We have used all five pentagons, so this face must be a hexagon:

But then we have a square face with a vertex of valence two, or else adding edges turns it into a degenerate face with too many edges, an impossibility.
Case 4
Suppose $F$ is a hexagon. By symmetry, the face on the other side cannot be a square or pentagon, so must also be a hexagon, like so:

There are only two open vertices.
An edge between them would create a square with the top edges, but also a triangle with the bottom two edges, which is forbidden.

Making a pentagon on the top instead, we have a square on the outside, but the open vertex remaining forces us to add edges, making a degenerate face with too many sides:

Finally, we can complete a hexagon on the top:

Then the four green edges are part of a face. If this face is a pentagon, then the top black edge is part of a digon (forbidden). If this face is a hexagon, then the top black edge is part of a triangle (forbidden).

Thus, by exhausting cases, the desired graph cannot exist.
There are some solutions which have four pentagons around a square.
The first has two hexagons, and 16 vertices.
The other one has six hexagons and 22 vertices.

As I explained in comments, from each of these examples you can get an infinite set of examples by taking the bitruncation of each graph, which is the truncation of its dual. This has the effect of replacing each $n$-gonal face by a smaller $n$-gon inside, surrounded by hexagons which appear around each vertex.
