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 suppose the best way to tackle is to row-reduce the matrix and see what happens. You have to eliminate the third and the first columns:
$$
\begin{pmatrix}
1 & a & 1 & 3\\
1 & 2 & 2 & 6 \\
1 & 5 & 3 & b
\end{pmatrix}
\to
\begin{pmatrix}
1 &   a & 1 & 3\\
0 & 2-a & 1 & 3 \\
0 & 5-a & 2 & b-3
\end{pmatrix}
\to
\begin{pmatrix}
1 & 2a-2 & 0 & 0\\
0 & 2-a  & 1 & 3 \\
0 & 1+a  & 0 & b-9
\end{pmatrix}
$$
so if $a=-1$ the last equation reads $0=b-9$, thus having $b=9$ you have $0=0$, which eliminates the equation, and now you have 2 equations in 3 unknowns, hence an infinite number of solutions. Thus (i) cannot be the answer.
The condition for no solutions would require that $a=-1$ (forcing the last equation to read $0=b-9$) and $b \ne 9$ (forcing it to be false).
UPDATE
In retrospect, as you say, if $a=-1$ then $\Delta = 0$, so there will not be a unique solution. However, if $b = 9$, the last column of $\Delta_z$ is $(3,6,9) = 3 \times (1,2,3)$, so exactly $3$ times the last column of $\Delta$, hence, $\Delta_z = 3\Delta = 0$, and the idea in your book is correct.
