You may first find the determinant of the coefficient matrix $A$ for the left hand side of the system of equations. You should find that it is $(a+2)(a+4)$. In other words when $a\neq-2,-4$, the system always has a unique solution $(x,y,z)$, regardless of the values of $b_1,b_2,b_3$.
When $a=-2$ or $-4$, the coefficient matrix is singular. So, the system can be inconsistent (i.e. insolvable) if $b_1,b_2,b_3$ have not right values. For $a=-2$, you can apply row operations to bring the system of equations into a row reduced echelon form. You will find that the bottom row of the LHS will become zero, while the RHS becomes a nonzero multiple of $b_2-b_1-b_3$. Hence the system is solvable if and only if $b_2=b_1+b_3$.
Alternatively, you may use column operations to show that the column space is the span of $(0,1,1)^T$ and $(1,1,0)^T$. Hence the system is solvable iff $(b_1,b_2,b_3)=p(0,1,1)+q(1,1,0)$ for some $p$ and $q$, i.e. iff $b_2=b_1+b_3$. The case $a=-4$ can be handled in a similar manner.