let me try to solve this problem in a practical way. define $U = span\{u_{1},u_{2},u_{3}\}$, corresponding to the values given in your question. then there exist $u = a u_{1} + b u_{2} + c u_{3}$ such that $v-u \bot U$ ($\mathbb{R}^{4}$ is an inner product space, and orthogonality between x and y is difined by < x,y> = 0).
so in the basis form of vector (i assume the basis used in your question statement is orthogonal already)
$<\begin{bmatrix} 1\\ 0\\ 1\\ 0\end{bmatrix} -a \begin{bmatrix} 0\\ 0\\ 1\\ 0\end{bmatrix} -b\begin{bmatrix} 1\\ 1\\ 1\\ -1\end{bmatrix} -c\begin{bmatrix} 1\\ 3\\ 1\\ 1\end{bmatrix} ,n> = 0 $
for any $n$ in span U. substitute $u_{1},u_{2},u_{3}$ into above equation. you will find a system of 3 equations with 3 unknowns. it should be solvable.
For example, $<\begin{bmatrix} 1-b-c\\ 3-b-3c\\ 1-a-b-c\\ 1+b-c\\ \end{bmatrix}, \begin{bmatrix} 0\\ 0\\ 1\\ 0\end{bmatrix} > =1-a-b-c=0 $
Eventually, $ \begin{matrix} 1-a-b-c =0\\ 2-a-4b-4c = 0\\ 2-a-4b-12c = 0 \end{matrix} $