What is the solution to the following system ? [ 3x+3y+6z=9 [ x+3y+2z=5 [ 3x+12y+12z=18 a - (3,1,0) b - (0,1,0) c - (2,1,0) d - (-2,1,0)

3x+3y+6z=9
x+3y+2z=5
3x+12y+12z=18

Step 1: use Eq 2 to solve for x
x+3y+2z=5
x= -3y-2z+5
Step 2: Solve for y unsing Eq 1
3x+3y+6z=9
3y= -3x-6z+9
y= -x-2z+3

Step 3: plug in x= -3y-2z+5 to solve for the value of y
y=-x-2z+3
y= (-1)(-3y -2z +5) - 2z+3
y= 3y+2z-5- 2z+3
-2y= -2
y=1

Step 4: plug y= 1 into the x= equation 
x= -3y-2z+5
x= -3(1)-2z+5
x= -2z+2
Step 5:Plug y= 1 and x= -2z+2 into the 3rd equation to solve for z
3x+12y+12z=18
12z= -3(-2z+2)-12(1)+ 18
12z= 6z - 6 - 12 +18
6z= -18+18
6z=0
z=0

Step 6: Plug z in to solve for x
x=-2z+2
x= (-2)(0)+2
x= 2

ANSWER:
x,y,z = 2,1,0
answer choice C

Ihave, but this question is gonna probably get deleted.