Having a row of $0$'s in the row-echelon form means that we were able to write the third row of $A$ as a linear combination of the second and first rows. As it so happens for square matrices, this is true precisely when we can write the columns as a linear combination of each other (that is, when the columns are not linearly independent). If you further reduce this to reduced row-echelon form, you get
$$\begin{bmatrix}
1 & 0 & -4/3 & 0\\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0
\end{bmatrix}$$
Because the third column lacks a pivot, $x_3$ is our free variable, which means that we can write $a_3$ as a linear combination of the other two columns.
There's a very good reason for focusing on the columns of a matrix. This comes out of using $A$ as a linear transformation, where the "column space" gives us the "range" of the function $f(\vec x) = A \vec x$.
So the question is to find the span of $v_{1} = \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}$, $v_{2} = \begin{bmatrix} 3 \\ 1 \\ 1 \\ \end{bmatrix}$, $v_{3} = \begin{bmatrix} 9 \\ 4 \\ -2 \\ \end{bmatrix}$, and $v_{4} = \begin{bmatrix} -7 \\ 3 \\ 1 \\ \end{bmatrix}$.
Before trying to solve this problem, it is important to know what span means. The first thing you need to know is where the vectors live. To figure this out, count the number of components in one of the vectors. In this case, there are 3 components in each vector, so these vectors live in $\mathbb{R}^{3}$. The first component corresponds to the $x$-axis, the second component corresponds to the $y$-axis, and the third component corresponds to the $z$-axis.
Now, if we take two linearly independent vectors, say $v_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ and $v_{2} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}$, then these vectors clearly live in $\mathbb{R}^{3}$, since they each have 3 components. Since they are linearly independent, and there are 2 of them, they span a plane in $\mathbb{R}^{3}$. This particular plane is isomorphic to $\mathbb{R}^{2}$, but it is not $\mathbb{R}^{2}$, because vectors in $\mathbb{R}^{2}$ all look like $\begin{bmatrix} x \\ y \\ \end{bmatrix}$ (i.e., they have 2 components only). A plane in $\mathbb{R}^{3}$ is clearly 2-dimensional since that is how we define a plane, but we do not describe it as $\mathbb{R}^{2}$. Instead, we simply say it is a plane in $\mathbb{R}^{3}$.
Here is an example of what I mean: span$\left \{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right \}$ is the $XY$-plane in $\mathbb{R}^{3}$, while span$\left \{\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}$ is the $XZ$-plane in $\mathbb{R}^{3}$, and span$\left \{\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right \}$ is the $YZ$-pane in $\mathbb{R}^{3}$. All three of the spans I just mentioned are isomorphic to $\mathbb{R}^{2}$, but they are distinct planes in $\mathbb{R}^{3}$, which is why we must describe them as planes in $\mathbb{R}^{3}$, rather than just as $\mathbb{R}^{2}$. When described as planes, they can be differentiated from each other, which is good because they are different planes.
Now, on to your question, as you correctly stated, to determine which vectors are linearly independent (in order to determine the dimension of the span), we put the vectors as columns in a matrix and reduce to RREF. So after doing that, we get that
$\begin{bmatrix} 1 & 3 & 9 & -7 \\ 0 & 1 & 4 & 3 \\ 2 & 1 & -2 & 1 \\ \end{bmatrix}$ reduces to $\begin{bmatrix} 1 & 0 & -3 & 0 \\ 0 & 1 & 4 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$
We see that there are three pivots! The number of pivots is the number of linearly independent columns, and the pivots correspond to the linearly independent columns, so the vectors $ \left \{ \begin{bmatrix} 1 \\ 0 \\ 2 \\ \end{bmatrix}, \begin{bmatrix} 3 \\ 1 \\ 1 \\ \end{bmatrix}, \begin{bmatrix} -7 \\ 3 \\ 1 \\ \end{bmatrix} \right \}$ (which are $v_{1}$, $v_{2}$, and $v_{4}$) are the linearly independent ones, and $v_{3}$ is linearly dependent on them. Since there are three linearly independent vectors, the span of all four vectors is equal to the span of the three linearly independent ones. Three linearly independent vectors span a subspace that is 3-dimensional. But these vectors live in $\mathbb{R}^{3}$, which is 3-dimensional itself, so their span must be equal to $\mathbb{R}^{3}$. If these vectors happened to live in $\mathbb{R}^{4}$, then their span would be a 3-dimensional subspace of $\mathbb{R}^{4}$.
So to answer your question, these four vectors could have spanned a 2-dimensional subspace of $\mathbb{R}^{3}$ if only two of the four were linearly independent. But we had three pivots in our matrix, so three linearly independent vectors, which meant the span of the four was equal to the span of the three linearly independent vectors (i.e., span$\{ v_{1}, v_{2}, v_{3}, v_{4} \} = $ span$\{ v_{1}, v_{2}, v_{4} \}$), and they spanned all of $\mathbb{R}^{3}$ in this case because they live in the 3-dimensional space and their span is 3-dimensional.
Best Answer
You will find the answer to this in most introductory linear algebra textbooks but a quick answer would be as follows: