Saudi Arabian Mathematical Competitions

Let $a, b, c$ be three distinct elements in $S$.
If we denote by $\mathcal{S}_i$, the set of subsets of $S$ of $i$ elements, $i=2,3$, we notice that
$$\sum_{\{x, y\} \in \mathcal{S}_2} x y = \frac{1}{3} \sum_{\{x, y, z\} \in \mathcal{S}_3} (x y + y z + z x) \in \mathbb{Q} .$$
If we denote by $\mathcal{S}_{x, i}$, the set of subsets of $S \setminus \{x\}$ of $i$ elements, $i=2,3$, for $x \in S$, we notice that
$$a c = \sum_{\{x, y\} \in \mathcal{S}_2} x y - \frac{1}{2} \left( \sum_{\{x, y, z\} \in \mathcal{S}_{a, 3}} (x y + y z + z x) + \sum_{\{x, y, z\} \in \mathcal{S}_{c, 3}} (x y + y z + z x) \right) .$$
Hence $a c \in \mathbb{Q}$. Similarly $b c \in \mathbb{Q}$. We deduce that $\frac{a}{b} = \frac{a c}{b c} \in \mathbb{Q}$.