Solve the following system of equations:
x + 3y − z = 2
x − 2y + 3z = 7
x + 2y − 5z = −21
(2, 3, 5)
(−2, 3, 5)
(2, −3, 5)
(2, 3, −5)

From the first equation: x = 2 - 3y + z . . . . . . . . (1)
From the second equation: x = 7 + 2y - 3z . . . . . . . . (2)
From the third equation: x = -21 - 2y + 5z . . . . . . . . (3)
Equating (1) and (2), we have:
2 - 3y + z = 7 + 2y - 3z
5y - 4z = -5 . . . . . . . . (4)
Equating (1) and (3), we have:
2 - 3y + z = -21 - 2y + 5z
y + 4z = 23 . . . . . . . . (5)
(4) + (5) gives: 6y = 18
y = 18/6 = 3
From (5), 3 + 4z = 23
4z = 23 - 3 = 20
z = 20/4 = 5
From (1), x = 2 - 3(3) + 5 = 2 - 9 + 5 = -2.

Therefoe, solution = (-2, 3, 5)