Answer
See answer below
Work Step by Step
Set the values to 0,
$av_1+bv_2+cv_3+dv_4=0$
we have:
$\begin{bmatrix}
1 & 1 & 1 | 0\\
1 & -1 & 1 | 0\\
1 & -3 & 1 |0\\
3 & 1 & 2| 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 1 & 1 | 0\\
0 & 2 & 0 | 0\\
0 & 4 & 0 |0\\
0 & 2 & 1| 0
\end{bmatrix}\approx \begin{bmatrix}
1 & 1 & 1 | 0\\
0 & 2 & 0 | 0\\
1 & 0 & 1 |0\\
0& 0 & -1| 0
\end{bmatrix}$
Since the first, second and fourth vetors are linearly independent, we have
$\begin{bmatrix}
1 & 1 & 1\\
0 & 2 & 0\\
0 & 0 & -1
\end{bmatrix}$
$rank (A_2)=3$
Hence, the set $\{ v_1, v_2, v_3 \}$ spans the same subspace of V as that spanned by the original set of vectors $\{ v_1, v_2, v_3, v_4 \}$