Solve the following system
2x+3x-z=1
3x+y+2z=12
x+2y-3z=-5
a) (3, 1, 2)
b) (-3, 1 , 2)
c) (3 , -1 , 2)
d) (3 , 1 , -2)

2x + 3y - z = 1 . . . . . . . . (1)
3x + y + 2z = 12 . . . . . . .(2)
x + 2y - 3z = -5 . . . . . . . .(3)
From (3), x = -2y + 3z - 5
Sustituting for x, in (1) and (2), we have
2(-2y + 3z - 5) + 3y - z = 1
-4y + 6z - 10 + 3y - z = 1
-y + 5y = 11 . . . . . . . . (4)
3(-2y + 3z - 5) + y + 2z = 12
-6y + 9z - 15 + y + 2z = 12
-5y + 11z = 27 . . . . . . .(5)
(4) * 5 = -5y + 25z = 55 . . . .(6)
(5) - (6) = -14z = -28
z = -28/-14 = 2
From (5), -5y + 11(2) = 27
-5y = 27 - 22 = 5
y = 5/-5 = -1
x = -2(-1) + 3(2) - 5 = 2 + 6 - 5 = 3

Solution = (3, -1, 2)

Your answer should be “b” i truly hope i could ! good luck (: