Answer:
The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4).
Let its vertices be A(x1,y1,z1), B(x2,y2,z2), C(x3,y3,z3) .
The mid point of AB is (1,5,-1).
x1 + x2 = 2 …..(1)
y1 + y2 = 10 …..(2)
z1 + z2 = -2 …..(3)
Mid point of AC is (2,3,4),
x1 + x3 = 4 …..(4)
y1 + y3 = 6 …..(5)
z1 + z3 = 8 …..(6)
Mid point of BC is (0,4,-2),
x2 + x3 = 0 …..(7)
y3 + y2 = 8 …..(8)
z3 + z2 = -4 …..(9)
Adding the equations 1,4 and 7, and divide it by 2,
x1 + x2 + x3 = 3
Subtracting 1, 4, 7 individually,
x1 = 3
x2 = -1
x3 = 1
Adding the equations 2,5 and 8, and divide it by 2,
y1 + y2 + y3 = 12
Subtracting 1, 4, 7 individually,
y1 = 4
y2 = 6
y3 = 2
Adding the equations 3,6 and 9, and divide it by 2,
z1 + z2 + z3 = 1
Subtracting 1, 4, 7 individually,
z1 = 5
z2 = -7
z3 = 3
The coordinates are A(3,4,5), B(-1,6,-7) and C(1,2, 3).