#{(2x+ay-3z=2),(x+y-z=1),(-x+by+z=3):}#
I'm using Gaussian elimination with smart choice of the first equation and substitution.
Add second equation to third and subtract from first two times
#{(2x-color(red)(2x)+ay-color(red)(2y)-3z-(color(red)(-2z))=2-color(red)(2)larr),(x+y-z=1rarr),
(color(red)(x)-x+by+color(red)(y)+z+(color(red)(-z))=3+color(red)(1)larr):}#
Combine like terms
#{(color(red)((a-2)y-z=0)),(x+y-z=1),(color(red)((b+1)y=4)):}#
We got #y#
#{((a-2)y-z=0),(x+y-z=1),(color(red)(y=4/(b+1))):}#
Assuming #b!=-1#. If #b=-1# then #0=4# we have contradiction and no solutions.
Substitute
#{((a-2)color(red)(4/(b+1))-z=0larr),(x+color(red)(4/(b+1))-z=1larr),(y=4/(b+1)rarr):}#
Move known terms to the right
#{(-z=-color(red)((4(a-2))/(b+1))),(x-z=1-color(red)(4/(b+1))),(y=4/(b+1)):}#
We got #z#
#{(color(red)(z=(4(a-2))/(b+1))),(x-z=1-4/(b+1)),(y=4/(b+1)):}#
Substitute
#{(z=(4(a-2))/(b+1)rarr),(x-color(red)((4(a-2))/(b+1))=1-4/(b+1)larr),(y=4/(b+1)):}#
Move known terms to the right
#{(z=(4(a-2))/(b+1)),(x=1-4/(b+1)+color(red)((4(a-2))/(b+1))),(y=4/(b+1)):}#
Simplify and we finally got it
#{(color(red)(x=1+(4(a-2)-4)/(b+1)=1+(4(a-3))/(b+1))),(y=4/(b+1)),(z=(4(a-2))/(b+1)):}#
