The vectors are $(1, 2, 3, 4)$, $(5, 6, 7, 8)$, $(9, 10, 11, 13)$ and $(15, 18, 21, 25)$.
I know the last vector is the sum of the other three, that are linear independent, so they span a 3-dimensional subspace of $\mathbb R^4$ . My problem is to determine this subspace.
For example, these two $\mathbb R^3$ vectors span a 2-dimensional subspace: $(1, 2, 1), (2, 0, -1)$. It is a plane and I can find the normal vector to it with the cross product of the vectors, being able to write the subspace $U = \{(x, y, z)\in \mathbb{R} | 2x-3y+4z=0\}$.
I was trying to write the subspace of the exercise similar to this of the plane, but was unable to develop the concept of a $\mathbb R^3$ subspace of $\mathbb R^4$.
Let us find a vector $\;v=(a,b,c,d)\in\Bbb R^4\;$ s.t. $\;v\perp U:=Span\{\,(1,2,3,4), (5,6,7,8), (9,10,11,13)\,\}\;$, and for this it is enough to find the solution to the homogeneous system determined by these three vectors:
$$\begin{pmatrix}1&2&3&4\\5&6&7&8\\9&10&11&13\end{pmatrix}\longrightarrow\begin{pmatrix}1&2&3&4\\0&-4&-8&-12\\0&-8&-16&-23\end{pmatrix}\longrightarrow\begin{pmatrix}1&2&3&4\\0&-4&-8&-12\\0&0&0&1\end{pmatrix} $$
and we thus get that the solution space of dimension $\;1\;$, which is spanned say by $\;(1,-2,1,0)\;$, from which you can write that $\;U=\left(Span\left\{(1,-2,1,0)\right\}\right)^\perp\;$
Another way: Take the vector product of the three lin. ind. vectors in $\;\Bbb R^4\;$ to get an orthogonal vector to them all. Puting $\;\ell\;$ as the fourth dimensional unit vector, we get:
$$\begin{vmatrix}i&j&k&\ell\\1&2&3&4\\5&6&7&8\\9&10&11&13\end{vmatrix}=(-4,8,-4,0)$$
and we get a scalar multiple of the first vector above...but calculating big determinants can be tough