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.
The Gaussian elimination that you performed showed only one thing: that the three original vectors are linearly independent
Because they are linearly independent, they form a basis for the space that they span. That space is not all of $\mathbb{R}^4$, but is a three dimensional subspace of $\mathbb{R}^4$. For that subspace, they form a basis.
But they are certainly not a basis for all of $\mathbb{R}^4$. All of $\mathbb{R}^4$ is $4$-dimensional: all of its bases have exactly $4$ (linearly independent) vectors.
Finally, that subspace is not $\mathbb{R}^3$. It's just a $3$-dimensional subspace of $\mathbb{R}^4$. The symbol $\mathbb{R}^3$ refers to "triplets of numbers". Here, we're are not dealing with triplets of numbers. We are still dealing with quadruplets of numbers (which, to repeat, all live in a $3$-dimensional subspace of $\mathbb{R}^4$).
Here's a link to the appropriate place in our Linear Algebra course: http://lem.ma/Dh (although you may need to start a little bit earlier to get used to our approach).
Best Answer
The notation $\operatorname{span}(X)$ should be used for a set $X\subseteq V$ that is contained in some vector space $V$. It is then defined as the intersection of all subspaces of $V$ containing $X$ or equivalently as the set of all linear combinations of elements from $X$.
Writing "$\operatorname{span}(A)$" when $A$ is a matrix, doesn't make any sense in terms of this definition. However, it might be an abuse of notation to denote the span of the set of columns of $A$, that is, the column space of $A$. This space is usually denoted $\operatorname{col}(A)$ though.
The question about a "minimum number of vectors in a set $X$ to guarantee $b\in\operatorname{span}(X)$" is unclear to me. For example in the plane $V=\mathbb R^2$ you can take the infinite set $$ X = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \end{bmatrix}, \begin{bmatrix} 3 \\ 0 \end{bmatrix}, \begin{bmatrix} 4 \\ 0 \end{bmatrix}, \dots \right\} $$ with span $$ \operatorname{span}(X) = \left\{ \begin{bmatrix} x \\ 0 \end{bmatrix} \,\middle|\, x\in\mathbb R\,\right\} $$ the does not contain $b=\begin{bmatrix}0 \\ 1\end{bmatrix}$ despite $X$ having infinitely many elements.