In short, row reduced echelon form(RREF) of a matrix $A$ is such that
i) Every leading entry is 1
ii) Any nonzero rows are above zero rows
iii) any leading entry is strictly to the right of any leading entries above that row
iv) any other entry in a column containing a leading entry is 0 except for the leading entry.
So it does not have to be put in augmented matrix $[A|b]$ to get a RRE form. You are comparing RRE form of matrix $A$ and $[A|b]$.
To see why the statement is true, suppose that you put the matrix $[A|b]$ into RRE form, so you have a matrix E. If E contains a leading entry in its last column, in terms of system of equations, what does it say? And what is the condition for E to not have any leading entry in last column?
Note: If RRE form of $[A|b]$ does contain a leading entry, then it is different from that of $A$. Also, note that RRE form of $[A|b]$ is m by n+1 whereas that of $A$ is m by n.
Solve:
$x+y=1$
$x+y=2$
Then we have
$\
A =
\left( {\begin{array}{cc}
1 & 1 \\
1 & 1
\end{array} } \right)
$
$\
b =
\left( {\begin{array}{cc}
1 \\
2
\end{array} } \right)
$
and $Ax=b$
If we turn A into RREF, we get
$\
E =
\left( {\begin{array}{cc}
1 & 1 \\
0 & 0
\end{array} } \right)
$
So A has rank 1
and if we put $[A|b]$ into RRE form, we get
$\
E' =
\left( {\begin{array}{cc}
1 & 1 & 0 \\
0 & 0 & 1
\end{array} } \right)
$
So augmented matrix has rank 2. Observe what last row says in terms of equations.
There are various ways to determine whether the first matrix can be represented as a linear combination of the others.
The simplest way could be to solve for $a,b,c\in \mathbb{R}$ such that the following is true:
$
\begin{bmatrix}
-12\\
12\\
-18\\
-8
\end{bmatrix}=a\begin{bmatrix}
3\\
-3\\
3\\
3
\end{bmatrix}+b
\begin{bmatrix}
3\\
-2\\
4\\
5
\end{bmatrix}+c
\begin{bmatrix}
6\\
-5\\
10\\
6
\end{bmatrix}$
This can be turned into a system of equations shown below:
$$3a+3b+6c=-12$$
$$-3a-2b-5c=12$$
$$3a+4b+10c=-18$$
$$3a+5b+6c=-8$$
I will leave it up to you to solve for such $a,b,c$.
Another method you can use to approach this problem is to check for linear dependence. You can do this as you said by row-reducing the 4x4 matrix formed by the 4 column vectors.
$
\begin{bmatrix}
-12&3&3&6\\
12&-3&-2&-5\\
-18&3&4&10\\
-8&3&5&6
\end{bmatrix}$
If the row reduction leads you to the identity matrix, then you have a matrix of full rank (none of the column vectors are linearly dependent). However, if you have arrive at a matrix which only has 3 or fewer leading 1s, then you have a matrix whose column vectors are linearly dependent.
Best Answer
There are three spaces commonly associated with a matrix: column space, row space and null space.
One solution to the system of linear equations $Ax=b$ is $x=\begin{bmatrix}\color{blue}{2} \\ \color{red}{3} \\ \color{green}{0} \end{bmatrix}$, as can be seen from the given row echelon form. This implies the solution $$\require{cancel}\underbrace{\begin{bmatrix}2 \\ -1 \\ 6\end{bmatrix}}_b=\color{blue}{2}\begin{bmatrix}1 \\ -2 \\ 0\end{bmatrix}+\color{red}{3}\begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}+\cancel{\color{green}{0}\begin{bmatrix} 5 \\ -6 \\ 8 \end{bmatrix}}$$ which could alternatively be found by inspection. This implies $b$ is in the column space of $A$.
We can also see that $(2,-1,6)$ is the negative of the second row of $A$. This implies $b$ is in the row space of $A$ (or, perhaps more formally, $b^T$ is not in the row space of $A$).
Converting $A$ to row echelon form $R$ gives $$ R=\begin{bmatrix} 1 & 0 & 5 \\ 0 & 1 & 4 \\ 0 & 0 & 0 \\ \end{bmatrix} $$ and $Rx=\mathbf{0}$ has the solution space $\{(-5t,-4t,t):t \in \mathbb{R}\}$ -- this is the null space of $A$. Since $b$ is not of the form $(-5t,-4t,t)$, we see that $b$ is not in the null space of $A$.