I don't understand one step in the proof of Theorem 6.3B in the book "Permutation Groups" by J.D.Dixon-M.Mortimer.
Let 00, 01, ..., 22 be the points of affine geometry $AG_2(3)$. Then there are exactly nine quadrangles containing points 00 and 01; they contain these additional points: $Q_1: 10,11$; $Q_2: 10,12$; $Q_3: 10,21$; $Q_4: 11,12$; $Q_5: 11,20$; $Q_6: 12,22$; $Q_7: 20,21$; $Q_8: 20,22$; $Q_9: 21,22$. Denote the set of these nine quadrangles $\cal Q(00,01)$.
Let $\cal S$ be a set of 18 quadrangles of $AG_2(3)$ such that each triangle of $AG_2(3)$ is in a unique quadrangle from $\cal S$. Since there are six triangles which contain the points 00 and 01 and each quadrangle contains four triangles it follows that $\cal S$ must contain exactly three of the quadrangles $Q_1$ - $Q_9$. It is now said that it leaves only three possibilities for $\cal S\cap \cal Q(00,01)$: $\{Q_1,Q_6,Q_7\}$, $\{Q_2,Q_5,Q_9\}$ or $\{Q_3,Q_4,Q_8\}$. I don't understand why there is not a possibility $\{Q_3, Q_5, Q_6\}$.