In any quadruple of the points, consider the point which has the power $k$ with respect to the circle passing through the other three, name these points good points. We claim that there are two quadruples with the same good point, and two other common points.
There are $\binom{6}{4} = 15$ quadruples, each has at least one good point, since there are six points in total, there exists a point $P$ which is a good point in at least $\lceil \frac{15}{6} \rceil = 3$ quadruples.
The other five points need to fill the remaining place in each of these quadruples, there are a total of $3 \times 3 = 9$ places, so there is a point $Q$ that is in at least $\lceil \frac{9}{5} \rceil = 2$ of the quadruples.

Consider these two quadruples containing $P$ and $Q$. There are a total of 4 places in these quadruples need to be filled with the other 4 points. The claim is directly proved if some point $R$ is a member of both of these quadruples. In other case, the remaining 4 points must appear exactly once in these 4 remaining places. So we have two quadruples
$$Q_1 = (P, Q, R, S), \ Q_2 = (P, Q, T, U).$$
Where $P, Q, R, S, T, U$ are all six points of the problem.
We said $P$ is a good point in at least 3 quadruples. Consider the third quadruple $Q_3$ containing $P$ as a good point. 3 remaining places must be filled with the other five points $Q, R, S, T$ or $U$. If both $R, S$ or $T, U$ are members of $Q_3$, then (respectively) $Q_1$ or $Q_2$ have three common points with $Q_3$. Otherwise, only one member of $\{R, S\}$, one member of $\{T, U\}$ along with $Q$ are members of $Q_3$, in this case, both $Q_1, Q_2$ have three common points with $Q_3$. Therefore the claim is proved in any case.
So we have found two quadruples with exactly three common points and also the same good point $P$. Without loss of generality, assume that these two quadruples are $(P, Q, R, S)$, $(P, Q, R, T)$. Let $\omega_1$ and $\omega_2$ be the circumcircles of triangles $QRS$ and $QRT$, respectively. We have
$$\mathcal{P}_{\omega_1}(P) = \mathcal{P}_{\omega_2}(P) = k,$$
(Where $\mathcal{P}_{\lambda}(X)$ is the power of point $X$ with respect to circle $\lambda$.) If $\omega_1 \neq \omega_2$, the above equation implies $P$ is a point on the radical axis of these two circles, that means $P$ lies on $QR$ which is impossible since there are no three points on a same line. So the only possible case is when $\omega_1 = \omega_2$, means points $Q, R, S$ and $T$ lie on a circle. Therefore by considering the quadruple $(Q, R, S, T)$, we obtain $k = 0$.
The rest of the problem is clear, $k = 0$ means between any quadruple, one point lies on the circle passing through the other three point, in other words, any four points are concyclic. Now fix three points $P, Q, R$, we get that any other point is concyclic with these three points, hence all six points lie on the circumcircle of triangle $PQR$. ■