I was struggling with this one, so I used software to compute the reduced row echelon forms (which are unique to each linear system) for each and found them to be equal to each other, thus I need that the first matrix could be reduced to the second. Now, $$\begin{pmatrix} 1&2&-1&-3\\ 3&5&k&-4\\9&(k+13)&6&6\end{pmatrix} -3R_1+R_2\to R_2 \begin{pmatrix} 1&2&-1&-3\\ 0&-1&(3+k)&5\\9&(k+13)&6&9\end{pmatrix}$$ $$9R_1 -R_3 \to R_3 \begin{pmatrix} 1&2&-1&-3\\ 0&-1&(3+k)&5\\0&(5-k)&-15&-36\end{pmatrix} $$ $$-R_2 \to R_2 \begin{pmatrix} 1&2&-1&-3\\ 0&1&(-3-k)&-5\\0&(5-k)&-15&-36\end{pmatrix}$$ $$(5-k)R_2-R_3 \to R_3 \begin{pmatrix} 1&2&-1&-3\\ 0&1&(-3-k)&-5\\0&0&(k^2-2k)&(5k+11)\end{pmatrix}.$$
Span is usually used for a set of vectors. The span of a set of vectors is the set of all linear combinations of these vectors.
So the span of $\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}$ would be the set of all linear combinations of them, which is $\mathbb{R}^2$. The span of $\{\begin{pmatrix}2\\0\end{pmatrix}, \begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}$ is also $\mathbb{R}^2$, although we don't need $\begin{pmatrix}2\\0\end{pmatrix}$ to be so.
So both these two sets are said to be the spanning sets of $\mathbb{R}^2$.
However, only the first set $\{\begin{pmatrix}1\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}$ is a basis of $\mathbb{R}^2$, because the $\begin{pmatrix}2\\0\end{pmatrix}$ makes the second set linearly dependent.
Also, the set $\{\begin{pmatrix}2\\0\end{pmatrix}, \begin{pmatrix}0\\1\end{pmatrix}\}$ can also be a basis for $\mathbb{R}^2$. Because its span is also $\mathbb{R}^2$ and it is linearly independent.
For another example, the span of the set $\{\begin{pmatrix}1\\1\end{pmatrix}\}$ is the set of all vectors in the form of $\begin{pmatrix}a\\a\end{pmatrix}$.
Best Answer
The standard method for this kind of problem (I would think) is to combine the those spanning vectors and look for a complete set of relations between them.
Concretely you have $\def\sp{\operatorname{span}}U=\sp(u_1,u_2)$ and $V=\sp(v_1,v_2)$, and you are looking for common linear combinations $au_1+bu_2=cv_1+dv_2$, for any scalars $a,b,c,d$. That equation can also be written $au_1+bu_2-cv_1-dv_2=0$, which is a linear relation between $u_1,u_2,v_1,v_2$ with coefficients $a,b,-c,-d$ (the minus signs for $c,d$ are annoying, but you can consider $-c,-d$ as new unknown scalars). You can find all linear relations by solving a linear system $A\cdot(a,b,-c,-d)^T=(0,0,0)^T$ where $A$ has $u_1,u_2,v_1,v_2$ as columns. Here concretely $$ \begin{pmatrix}-3&-2&-5&-4\\-1&2&-1&2\\2&-1&0&-3\end{pmatrix} \cdot\begin{pmatrix}a\\b\\-c\\-d\end{pmatrix} =\begin{pmatrix}0\\0\\0\end{pmatrix}. $$ Solving this leaves one free parameter in the general solution, which I choose to be$~c$, and the general solution can be given as $(a,b,-c,-d)=c(1,-1,-1,1)$. This means that $u_1-u_2-v_1+v_2=0$ and this is essentially the only linear relation between those vectors (any other relation is a scalar multiple of it). The $u_1-u_2=v_1-v_2$ should be a common linear combination of $u_1,u_2$ and of $v_1,v_2$; indeed both sides check out to be $[-1,-3,3]$. Moreover all other common linear combination of $u_1,u_2$ and of $v_1,v_2$ are multiples of this vector, so $\{[-1,-3,3]\}$ is a basis of $U\cap V$.
More generally, the general solution of the homogeneous linear system $A\cdot\vec x=0$ can be written as a linear combination of certain specific solutions (with the coefficients being freely chosen parameters); those solutions (in the example there was only one) form a basis of $\ker(A)$. For each of those specific solutions one can take the first few coordinates and form the corresponding linear combination of the basis vectors of$~U$; this will also be minus the linear combination with the remaining coordinates of the basis vectors of$~U$, hence give a vector of $U\cap V$. Moreover, after running through the basis of $\ker(A)$, one is guaranteed to have obtained a basis of $U\cap V$ (since all vectors in $U\cap V$ must correspond to some common linear combination).