how to find rapidly the coordinates of Q:
since Q is the center of gravity of the triangle ABC, so we have the following vector relationship
vecQA +vecQB +vecQC =vec0
vecQA=(x-3, y+2)
vecQB=(x-1, y+5)
vecQC=(x-7, y+5)
vec0=(0, 0)
so, vecQA +vecQB +vecQC =vec0 is equivalent to
x-3 +x-1+x-7 =0, and y+2+y+5+y+5=0 so 3x-11=0 implies x=11/3
and 3y+12=0 implies y=-12/3
finally the the coordinates of point Q is (11/3, -4)