The midpoints of the sides of a triangle are (1, 5, -1), (0, 4, -2) and (2, 3, 4).

Let its vertices be A(x_1,y₁,z₁), B(x₂,y₂,z₂), C(x₃,y₃,z₃) .

The mid point of AB is (1,5,-1), therefore

x₁ +x₂ = 2………..eq.1

y₁ + y₂ = 10……eq.2

z₁ + z₂ = -2……..eq.3

Mid point of AC is (2,3,4), therefore

x₁ +x₃ = 4………..eq.4

y₁ + y₃ = 6……eq.5

z₁ + z₃ = 8……..eq.6

Mid point of BC is (0,4,-2), therefore

x₂ +x₃ = 0………..eq.7

y₃ + y₂ = 8……eq.8

z₃ + z₂ = -4……..eq.9

now, adding the equations 1,4 and 7, and divide it by two we get,

x₁ + x₂ + x₃ = 3

now subtracting 1, 4, 7 individually, we get

x₁ = 3, x₂ = -1 and x₃ = 1

now, adding the equations 2,5 and 8, and divide it by two we get,

y₁ + y₂ + y₃ = 12

now subtracting 1, 4, 7 individually, we get

y₁ = 4, y₂ = 6 and y₃ = 2

now, adding the equations 3,6 and 9, and divide it by two we get,

z₁ + z₂ + z₃ = 1

now subtracting 1, 4, 7 individually, we get

z₁ = 5, z₂ = -7 and z₃ = 3

therefore, the coordinates are A(3,4,5), B(-1,6,-7) and C(1,2,3).