Lene them up vertically so that the letters are in 
separate columns:

 x - 2y + bz = 3
ax      + 2z = 2
5x + 2y      = 1

Since y is eliminated from the middle equation,
let's eliminate y from the 1st and 3rd equations.
We can do that simply by adding them

 x - 2y + bz = 3
5x + 2y      = 1
----------------
6x      + bz = 4

Now we have this system:

     ax + 2z = 2
     6x + bz = 4

Let's eliminate either x or z.  I'll pick z
to eliminate:

Multiply the first by b and the second by -2

     abx + 2bz = 2b
    -12x - 2bz = -8
  ----------------
(ab-12)x       = 2b-8

There are three cases:
1. The left side is 0 and the right side is not 0. There is no solution.
2. The left side is not 0. There can only be on value of x, and hence one value
   for y and z as well.
3. Both sides are 0. Then x can be any value and there can be infinitely many
   solutions. 

The coefficient of x is 0 when and only when ab-12=0 or ab=-12
The right side is 0 when and only when 2b-8=0, or b=4.


1. If the right side is NOT 0, then there can be no solution,
because a left side of 0 cannot equal to a right side which
is something other than 0.


2. If the coefficient of x is not 0, there is a unique solution,
for then  which will have a single value
for the denominator will not be 0.
So there is a unique solution if  ab ≠ 12


3.  There are infinitely many solutions if both the coefficient of
x and the right side are both 0.  That's because x can be any 
number whatever and 0x = 0 will always be true.  That will be
when ab = 12 and b = 4.

--------------------
Answer:

1. There are no solutions if  ab = 12 and b ≠ 4
2. There is one unique solution if  ab ≠ 12 
3. There are infinitely many solutions if ab = 12 and b = 4

Edwin