Let $P$ and $Q$ be given fulfilling the conditions of the problem. Considering the circumcircle $k$ of $APQ$, we note that the centroid $S$ of $APQ$ must lie on $k$, since both $\angle PAQ = 60^\circ$ and $\angle PSQ = 120^\circ$ hold. Since $|SQ| = |SP|$, the arcs $SQ$ and $SP$ are of equal length, and we therefore have $\angle SAP = \angle SAQ = 30^\circ$. All centroids $S$ therefore lie on the angle bisector $w_{\alpha}$ of $\angle BAC$. The most extreme positions of $S$ are assumed when $P$ coincides with $A$ or the mid-point $M_{AB}$ of $AB$. In the latter case, $S$ is also the centroid, i.e. the mid-point, of $ABC$. In the former, $S$ is the centroid of the triangle $AM_{AB}M_{AC}$ (where $M_{AC}$ is the mid-point of $AC$). The set of all centroids of triangles $PQR$ fulfilling all requirements is therefore the middle third of the bisector $w_{\alpha}$ (which is also the altitude in $ABC$).