System of equations has no solution 
If the system of linear equations,
  \begin{cases}
x &+ ay &+ z  &= 3\\
x &+ 2y &+ 2z &= 6\\
x &+ 5y &+ 3z &= b\\
\end{cases}
  has no solution, then:
  
  
*
  
*$a=-1,b=9$
  
*$a=-1,b \ne 9$
  
*$a\ne-1,b = 9$
  
*$a=1,b \ne 9$

I really fail to understand why the answer cannot be option (1). 
Here's my method: 
For $a = -1$, the coefficient determinant determinant is 0. And if we create $\Delta_z$, the determinant in which coefficients of $z$ are replaced by $3,6,b$, (b= 9), that determinant is non zero. 
My book says that if coefficient determinant is $0$ and at least one of $\Delta_y, \Delta_z, \Delta_x$  is non $0$ then the system has no solution. 
But the answer given is $(2)$. What's wrong with my method? What's the best way to tackle such problems? 

Rewrite the system as:
$$
AX=B
$$
where $X=(x,y,z)^t$, $B=(3,6,b)^t$ and
$$A = \begin{bmatrix}
    1 & a & 1 \\
    1 & 2 & 2 \\
    1 & 5 & 3
    \end{bmatrix}$$
First you must have $det(A)=0$ which after calculations give $a=-1$.
So the system becomes:
$$A = \begin{bmatrix}
    1 & -1 & 1 \\
    1 & 2 & 2 \\
    1 & 5 & 3
    \end{bmatrix}$$
Now, row 3 becomes all zero when $r_3\rightarrow r_3-2r_2+r_1$. Applying this to the augmented matrix $[A|B]$ gives the $(3,4)$ entry to be:
$$
b-9
$$
It is required that this entry is nonzero in order to have an inconsistent matrix.
Thus, $a=-1, b\neq 9$.
