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,
$$
\begin{matrix}
1-a-b-c =0\\
2-a-4b-4c = 0\\
2-a-4b-12c = 0
\end{matrix}
$$