Finding The Basis Of An Intersection Of Two Subspaces 
Let $$U=\operatorname{Span}\{v_1,v_2,v_3\}, V=\operatorname{Span}\{v_4,v_5,v_6\}$$ 
Where $$v_1=(1,28,2,39),v_2=(2,28,2,39),v_3=(-1,28,2,39)\\v_4=(0,8,0,11),v_5=(0,31,1,43),v_6=(0,-3,0,-4)$$
Find a basis for $U\cap V$

I have forgot the algorithm for finding intersection of subspaces, but in the exercise I was given an algorithm that I have yet met and would like to understand, how and why it works.
The first step is to find an homogeneous system s.t the subspace is the solution set (Null space).
To do so for $U$ we look at $$\left(\begin{array}{ccc|c}  
 1 & 2 & -1 &x\\  
 28 & 28& 28&y\\
 2 & 2 & 2& z\\
 39& 39&39& w  \\ 
\end{array}\right)\sim \left(\begin{array}{ccc|c}  
 1 & 2 & -1 &x\\  
 1 & 1& 1& \frac{y}{28}\\
 0 & 0 & 0& \frac{z}{2}-\frac{y}{28}\\
 0& 0&0& \frac{w}{39}-\frac{y}{28}  \\ 
\end{array}\right)$$
So the matrix that $U$ is her solution set is $$ \begin{pmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39} \end{pmatrix}$$
Doing the same with $V$ we get
$$\left(\begin{array}{ccc|c}  
 0 & 0 & 0 &x\\  
 8& 31& -3&y\\
 0 & 1 & 0& z\\
 11& 43&-4& w  \\ 
\end{array}\right)\sim \left(\begin{array}{ccc|c}  
 8 & 31 & -3 &y\\  
 0 & 1 & 0& z\\
0 & 0& 1& 8w-11y-3z\\
 0& 0&0& x  \\ 
\end{array}\right)$$
So the matrix that $V$ is her solution set is $$ \begin{pmatrix} 1&0&0&0\end{pmatrix}$$
To find the intersection we solution for  $$\left(\begin{array}{cccc|c}  
 0 & -\frac{1}{28} & \frac{1}{2} &0 &0\\  
 0 & -\frac{1}{28}& &\frac{1}{39}& 0\\
 1 & 0 & 0& 0 &0\\
 \end{array}\right) $$
Which is $$ \begin{pmatrix} 0&28&2&39\end{pmatrix}$$
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?
b. Why looking at the null space of the vectors (why we put the vectors in horizontal and not in vertical) that we found give us the basis of intersection?
C. Is there another way to find the basis of intersection? 
 A: For the point b, remember that the Null Space of the matrix V = empty set, because they are independent. What you found is the Null Space of the Transpose(V). Same thing about U. But for the intersection you must to swap rows and columns.
A: 
a. Why does putting the vectors in columns and $x,y,z,w$ in the $b$ vector give us a system which the vectors are their solutions?

Call $u=\begin{bmatrix} 1 & 2 & -1\\  
 28 & 28& 28\\
 2 & 2 & 2\\
 39& 39&39\end{bmatrix}$ and $\xi=\begin{bmatrix}x\\y\\z\\w\end{bmatrix}$.
Then $(x,y,z,w)\in U$ if and only if $\operatorname{rank}(u|\xi)=\operatorname{rank}(u)$.
The matrix $u$ is simplified by row operations which are performed with a multiplication on the left by a invertible matrix $a$, thus
$$a=\begin{bmatrix}
1&0&0&0\\0&\frac 1{28}&0&0\\0&-\frac 1{28}&\frac 12&0\\0&-\frac 1{28}&0&\frac 1{39}
\end{bmatrix}\quad
au=\begin{bmatrix} 1 & 2 & -1\\  
 1 & 1& 1\\
 0 & 0 & 0\\
 0& 0&0
\end{bmatrix}\quad a\xi=\begin{bmatrix}x\\\frac y{28}\\\frac z2-\frac y{28}\\\frac w{39}-\frac y{28}\end{bmatrix}$$
But $\operatorname{rank}(u|\xi)=\operatorname{rank}(u)$ if and only if $\operatorname{rank}(au|a\xi)=\operatorname{rank}(au)$.
This is equivalent to the system
$$\left\{\begin{array}{c}
\frac z2-\frac y{28}=0\\\frac w{39}-\frac y{28}=0
\end{array}\right.\iff\begin{bmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\end{bmatrix}\begin{bmatrix}x\\y\\z\\w\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\end{bmatrix}$$
hence $U$ is the kernel of your set solution
\begin{bmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\end{bmatrix}
Concluding: you put the generators of $U$ and $x,y,z,w$ in column in order to have
$$(x,y,z,w)\in U\iff\operatorname{rank}(u|\xi)=\operatorname{rank}(u)$$
and perform row operations; the kernel of the set solution gives your subspace $U$.
Thus we write our subspaces $U,V$ as kernel of two matrix.

b. Why looking at the null space of the vectors (why we put the
  vectors in horizontal and not in vertical) that we found give us the
  basis of intersection?

For matrix $A,B$ with the same number of columns, we have
$$\ker(A)\cap\ker(B)=\ker\begin{pmatrix}A\\B\end{pmatrix}$$
hence from
$$U=\ker
\begin{bmatrix} 0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\end{bmatrix}\quad V=\ker\begin{bmatrix}1&0&0&0\end{bmatrix}$$
we get
$$U\cap V=\ker\begin{bmatrix}0&-\frac{1}{28}&\frac{1}{2}&0\\ 0&-\frac{1}{28}&0&\frac{1}{39}\\1&0&0&0\end{bmatrix}$$
