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? 

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.
