The given equations are
x + y + z = 2
2x + 2y + 2z = 4
- 3x - 3y - 3z = - 6
From the first equation,
x = 2 - y - z
We would substitute x = 2 - y - z into the second and third equations. We have
2(2 - y - z) + 2y + 2z = 4
4 - 2y - 2z + 2y + 2z = 4
- 2y + 2y - 2z + 2z = 4 - 4
0 = 0
- 3(2 - y - z) - 3y - 3z = - 6
- 6 + 3y + 3z - 3y - 3z = - 6
3y - 3y + 3z - 3z = - 6 + 6
0 = 0
The equations would form the same line.
Thus, the correct answer is
B. Infinite solutions