Mutually disjoint triangles in planar cubic graph Let $G$ be a connected, planar graph for which every vertex has degree $3$ . Moreover, suppose there is a face of length at least $12$ . Is it possible to construct an example of such $G $ whose only odd faces are six triangles, and for which no two such triangles share a common vertex? (For my purposes I may assume there are no faces of length $1$ or $2$, in which case there must be an odd face of length at least 3. By the handshake lemma, there are an even number of, hence at least 2, such odd faces.)

Simplest answer: triangular prism.

Niftier solution: the chamfer of any cubic polyhedral having only triangular odd faces. (Chamfering) is a process where you replace each edge with a hexagon.)

Edit, if precisely 6 triangles are allowed, you want a chamfered cuboctohedron, or any further chamfering.

Edit 2: Start with a 12-gon, C. For 6 disjoint edges in C, create a triangle with a new vertex. Then, pair up triangles and create edges between their newest vertices.