Let us count the triples $(x, y, z)$ of positive integers such that $x + y + z = n$, where $n \ge 3$ is a given integer. One can assign to $x$ any value from $1$ to $n-2$. If $x=1$, then, taking into account that $z \ge 1$, $y$ can take $n-2$ values: from $1$ to $n-2$. If $x=2$, then $y$ can take $n-3$ values, ..., if $x=n-2$, then $y$ can take only the value $1$. This shows that the number of triples of type $n$ is
$$s(n) = 1 + 2 + 3 + \dots + (n-3) + (n-2) = \frac{(n-2)(n-1)}{2}.$$

a) If there exists $n$ such that $s(n) = 14$, then $(n-2)(n-1) = 28$, which is impossible: if $n \le 6$, then $(n-2)(n-1) \le 20$ and if $n \ge 7$, then $(n-2)(n-1) \ge 30$.