 Solution

As established above, $Q_3^{\prime }\cong K_{4,4}$. Select any three vertices in each parity class; all nine edges between the two triples are present, giving a $K_{3,3}$ subgraph. A planar graph has only planar subgraphs, whereas $K_{3,3}$ is nonplanar by the permitted result. Therefore $\boxed{Q_3^{\prime }\text{ is nonplanar}}$.