This is a mean problem because it would take lots
of solving and switching variables around to find
the solution, in all the different forms in the 
choices.

So what I would do in a case like this is pick a 
random value for x in the first 3 answers and for 
y in the 4th answer, and try to eliminate the wrong 
ones:

A.) (­-x, x + 2, 0)
B.) (x, x ­- 3, 0)
C.) (x + 2, x, 0)
D.) (0, y, y + 4) 

Pick 1 for x and y, and you get:

A.) (­-1, 3, 0)
B.) (1, -2, 0)
C.) (3, 1, 0)
D.) (0, 1, 5)

2x ­- 2y ­- z = 6 
-x + y + 3z = -­3 
3x ­- 3y + 2z = 9

try A.)   Substitute (x,y,z) =(-1,3,0)
in the first equation:

2(-1) ­- 2(3) ­- (-1) = 6
             -2-6+1 = 6
                 -7 = 6

That doesn't check in the first equation.
So (A) is eliminated

------------

try B.)   (x,y,z) =(1,-2,0)
in the first equation:

2(1) ­- 2(-2) ­- (0) = 6
             2+4-0 = 6
                 6 = 6

That checks in the first equation.  So (B) might be
the correct choice.  But let's check (C) and (D)
before we make the decision that (B) is the correct.
For it might not check in the other two equations.

    
------------

try C.)   (x,y,z) =(3,1,0)
in the first equation:

2(3) ­- 2(1) ­- (0) = 6
             6-2-0 = 6
                 4 = 6

That doesn't check in the first equation.
So C.) is eliminated
 
------------


try D.) (x,y,z) =(0,1,5)
in the first equation:

2(0) ­- 2(1) ­- (5) = 6
             0-2-5 = 6
                -7 = 6

That doesn't check in the first equation.
So D.) is eliminated

So the correct choice can only be B.)

Some teachers may frown on this method.  But
it is often the easiest way when you have a
multiple choice test like this, and it would
take a long time to solve in every form.

Edwin