Coefficient matrix determinant: Δ=1(15−2a)−1(6−a)+1(4−5)=8−a.
For a=8, Δ=0, so unique solution is not possible.
With a=8: R2−2R1 gives 3y+6z=24⇒y+2z=8.
R3−R1 gives y+2z=b−6.
For consistency: b−6=8⇒b=14.
When a=8,b=14: the system has infinitely many solutions.