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.
For the row basis, the non-zero rows in the RREF forms the basis. This is due to elementary row operations does not change the row space and also the non-zero rows are linearly independent.
Dimension of column space is equal to the number of columns with a pivot. It is known that the dimension of row space is equal to the dimension of column space. To see this, just notice that the number of pivot is equal to number of non-zero rows in the RREF.
You might find rank-nullity theorem helpful, that is
$$rank(A)+ nullity(A) = \text{number of columns}$$
Nullity is the dimension of null space.
Best Answer
You can solve it with matrices, sort of.
To check linear independence of a set $\{f_1, f_2, f_3\}$ of twice differentiable functions on $\mathbb{R}$, you can calculate the Wronskian, the determinant defined as:$$W(f_1, f_2, f_3)(t) = \begin{vmatrix} f_1(t) & f_2(t) & f_3(t) \\ f_1'(t) & f_2'(t) & f_3'(t)\\ f_1''(t) & f_2''(t) & f_3''(t)\end{vmatrix}$$
Notice that if your functions satisfy $\alpha f_1 + \beta f_2 + \gamma f_3 = 0$ on $\mathbb{R}$, then linearity of the derivative also implies $$\alpha f_1' + \beta f_2' + \gamma f_3' = 0$$ $$\alpha f_1'' + \beta f_2'' + \gamma f_3'' = 0$$
so $W(f_1, f_2, f_3)(t)$ will be zero on $\mathbb{R}$, since the matrix columns will be dependent.
Thus, if $W(f_1, f_2, f_3)$ is not identically zero, we can conclude that $\{f_1, f_2, f_3\}$ are linearly independent. However, if $W(f_1, f_2, f_3) = 0$ you cannot outright conclude that $\{f_1, f_2, f_3\}$ is linearly dependent.
In your case, we have:
$$W(e^{2t}, t^2, t)(t) = \begin{vmatrix} e^{2t} & t^2 & t \\ 2e^{2t} & 2t & 1\\ 4e^{2t} & 2 & 0\end{vmatrix} = -2 e^{2 t} (2t^2 - 2 t + 1) \ne 0$$
for all $t \in \mathbb{R}$ so $\{e^{2t}, t^2, t\}$ is certainly linearly independent, and hence a basis for your space.