Answer
See answer below
Work Step by Step
We can see that $0v_1+0v_3+0v_4=0=v_2$
Set the values to 0, we have:
$3x+y-z=0$
$x+2y+2z=0$
$5x-y+3z=0$
which is equal to: $\begin{bmatrix}
3 & 1 & -1\\
1 & 2 & 2\\
5 & -1 & 3
\end{bmatrix}$
$\det (A)=3\begin{vmatrix}
2 & 2\\
-1 & 3
\end{vmatrix}-\begin{vmatrix}
1 & 2 \\
5 & 3
\end{vmatrix}-\begin{vmatrix}
1 & 2 \\
5 & -1
\end{vmatrix}$
$\det (A)=3.[6-(-2)]-(3-10)-(-1-10)=42 \ne 0$
Hence, the set $\{ v_1, v_2, v_3 \}$ is linearly independent.