The medians of a triangle meet at a point known at the center of the triangle. Reversing the problem we may ask a relevant question:
The answer to the question is in positive, and the triangle can be constructed is as follows.
Pick point A on α, find points B on β and C on γ such that AB is bisected by γ and AC is bisected by β. There is a simple way to achieve that goal. (We already used this construction in finding a parallelogram cross-section of a pyramid.)
In ΔABC, β and γ house two of the medians BB' and CC'. The third median AA' meets them at the center of the triangle and lies, therefore, on α, implying that α bisects BC.
Obvioiusly, the construction is not unique, what is unique is the shape of the triangles - they all are similar.