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? 

I would do it this way, using row reduction. The criterion for a non-homogeneous linear system to have solutions is that the matrix of the linear system (l.h.s.) and the augmented matrix have the same rank.
From this criterion we deduce  readily that if this system has no solution, the matrix of the l.h.s. cannot have rank $3$, hence its determinant, which is equal to $-(a+1)$ is zero, whence $a=-1$. Computing the determinant by row reduction shows this matrix has rank $2$.
Now, if $a=-1$, the augmented matrix reduces as follows:
$$\begin{bmatrix}1&-1&1&3\\1&2&2&6\\1&5&3&b\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&-1&1&3\\0&3&1&3\\0&3&1&b-6\end{bmatrix}\rightsquigarrow
\begin{bmatrix}1&-1&1&3\\0&3&1&3\\0&0&0&b-9\end{bmatrix} $$
Thus, for $a=-1$, the augmented matrix has rank $3$ if and only if $b-9\ne 0$.
