Note, hectictar that the centroid is given by [ (x1 + x2 + x3)/ 3, (y1 + y2 + y3)/3 ]
Let B= (0,0) A = (0,3) and C = (4,0)....so
The centroid is [ (0 + 0 + 4)/3 , ( 0 + 3 + 0)/3] = [ 4/3, 1]
And as you pointed out, triangles ABG, BCG, and CAG, are all equal
So...area of ABG = (1/2)BH = (1/2)(AB)(GR) = (1/2)(3)(4/3) = 2 units^2
And the area of BCG = (1/2)(BC)(GP) = (1/2)(4)(1) = 2 units^2
And, by default, the area of the remaining triangle must be 2 units^2
So.....GQ kinda' "falls out" as
2 = (1/2)(AC)(GQ) = (1/2)(5)(GQ) = (5/2)GQ
So (5/2)(GQ) = 2
GQ = (2/5)(2) = 4/5
So......GR = 4/3 , GP = 1 and GQ = 4/5
So....your heights were correct.....
And their sum is = 47/15......just as you found...!!!