Consider the vectors (in $\mathbb{R}^3$):
$$ v_1=\begin{pmatrix}6\\1\\-3\end{pmatrix},~ v_2=\begin{pmatrix}-3\\3\\2\end{pmatrix}\text{ and } w=\begin{pmatrix}3\\4\\1\end{pmatrix} $$
is $w$ in $\text{span}(v_1,v_2)$?
I began setting it up to get it in reduced row echelon form but am not quite sure what that gives me. I am getting $x_1=-11$ and $x_2=5$ when doing that but I don't know if that proves anything or not.
Sorry about the formatting.
On
If $w$ is in span($v_1,v_2$), then we can express $w$ as the linear combination of $v_1$ and $v_2$, that is $$w = \lambda v_1+\mu v_2 \implies \begin{pmatrix}6\\1\\-3\end{pmatrix} = \lambda\begin{pmatrix}-3\\3\\2\end{pmatrix}\text{+ } \mu\begin{pmatrix}3\\4\\1\end{pmatrix},\ \lambda,\mu \in \mathbb{R}$$
From this equation, we have the following equation system: $$-\lambda+\mu = 2\ (I)$$ $$3\lambda+4\mu = 1\ (II)$$ $$2\lambda+\mu = -3\ (III)$$
Adding up $(II)$ and $(III)$, we have $$5\lambda+5\mu = -2 \implies \lambda+\mu = -\frac{2}{5}\ (IV)$$
Now, adding up $(I)$ and $(IV)$, we have $$2\mu = \frac{8}{5} \implies \mu = \frac{4}{5} \implies \lambda = -\frac{6}{5}$$
However, notice that these two values of $\lambda$ and $\mu$ doesn't satisfy $(II)$ and $(III)$ so we can say that we can't express $w$ as linear combination of $v_1$ and $v_2$. Therefore, $w$ is not in span($v_1,v_2$).
On
We can also find the span first, then check whether $w$ satisfies the equation of that subspace or not.
First, we know that span($v_1,v_2$) is the subspace which includes all the vectors that can be obtained by the linear combination of $v_1$ and $v_2$. Therefore, we have the equation $$\begin{pmatrix}x\\y\\z\end{pmatrix} =\lambda\begin{pmatrix}6\\1\\-3\end{pmatrix}+ \mu\begin{pmatrix}-3\\3\\2\end{pmatrix} = \begin{pmatrix}6\lambda-3\mu\\\lambda+3\mu\\-3\lambda+2\mu\end{pmatrix} ,\ \lambda,\mu \in \mathbb{R}$$
From here, we can say that $-11x+3y-21z = 0$ so we can express span as $$\text{span}(v_1,v_2) = \bigg\{(x,y,z)\ \big|\ z = \frac{-11x+3y}{21}\bigg\}$$ which is an equation for a plane in $3D$-space.
Now, it is easy to see that $x = 3$, $y = 4$, $z = 1$ doesn't satisfy the equation of the plane therefore $w$ is not on that plane, hence not in the $\text{span}(v_1,v_2)$.
On
Consider the matrix
\begin{bmatrix}6&-3&3\\1&3&4\\-3&2&1 \end{bmatrix}
We can reduce this matrix to row reduced echelon form by the following steps:
$\to$
\begin{bmatrix}3&-1&1\\0&-10&-11\\0&1&2 \end{bmatrix} $\to$
\begin{bmatrix}3&-1&1\\0&1&\frac{11}{10}\\0&1&2 \end{bmatrix} $\to$
\begin{bmatrix}3&0&3\\0&1&\frac{11}{10}\\0&1&2 \end{bmatrix} $\to$
\begin{bmatrix}1&0&1\\0&1&\frac{11}{10}\\0&0&\frac{-9}{10} \end{bmatrix} $\to$
\begin{bmatrix}1&0&0\\0&1&0\\0&0&1 \end{bmatrix}
Notice that in RREF, this matrix reduces to the identity matrix. This means our original matrix is rank 3 (full rank). This implies that these three vectors are linearly independent and hence cannot be written as a linear combination of the other vectors.
So to answer your question, $w$ is not in the $span$ of $v_1$ and $v_2$.
Hope this helps
The result depends upon the number of pivots you have in the RREF.
Notably
In this case
$$\begin{bmatrix}6&1&-3\\-3&3&2\\3&4&1 \end{bmatrix} \to\begin{bmatrix}6&1&-3\\-3&3&2\\0&7&3 \end{bmatrix} \to\begin{bmatrix}6&1&-3\\0&7&1\\0&7&3 \end{bmatrix} \to\begin{bmatrix}6&1&-3\\0&7&1\\0&0&2 \end{bmatrix} $$
Therefore
$$w\not \in span(v_1,v_2)$$
What is in the span is
$$u =\begin{pmatrix}3\\4\\-1\end{pmatrix}\quad u=v_1+v_2$$