Answer: (1, 3)
Let the vertices be \(A(x_1,y_1), B(x_2,y_2), C(x_3,y_3)\) with D, E, F the midpoints of the sides. Adding the three midpoint equations for x:
\[\frac{(x_1+x_2)+(x_2+x_3)+(x_3+x_1)}{2}=-7+8+2\]
\[x_1+x_2+x_3=3\]
Similarly \(y_1+y_2+y_3=6+5-2=9\).
Centroid \(=\left(\dfrac33,\dfrac93\right)=(1,3)\). (The triangle and the triangle of its midpoints have the same centroid.)