Answer: 7 sets.
Let $N$ be the required number of 3-element sets satisfying the problem condition. Suppose that $N \ge 8$. Let $M = \{x, y, z\}$ be one of these sets. Since the number of the remaining sets is greater than or equal to 7 and each such set has exactly one common element with $M$, we see that there exists an element of $M$ (say, $x$) that belongs to at least $[7/3] = 3$ sets. Let the sets $M_1, M_2, M_3$ contain $x$. By condition, there is not an element that belongs to all sets, so there exists a set (say, $M_0$) such that there is an element from $M$ (say, $y$), which belongs to $M_0$. It follows that $x \notin M_0$. Since any two of the sets $M_1, M_2, M_3$ and $M$ have not common elements except for $x$ (and $x \notin M_0$), so all four elements of the intersection of $M_0$ with $M_1, M_2, M_3$, $M$ are distinct. Therefore $M_0$ consists of at least four elements, a contradiction. Thus, $N \le 7$.

It remains to show the example for $N = 7$. The required sets (see the Fig.) are $\{A, K, B\}$, $\{A, P, M\}$, $\{A, N, C\}$, $\{B, M, C\}$, $\{B, P, N\}$, $\{C, P, K\}$, $\{K, M, N\}$.