Linear dependence of 3 vectors in $\mathbb{R}^4$ Let $a,b,c \in \mathbb{R},$ $\vec{v_1}=\begin{pmatrix}1\\4\\1\\-2 \end{pmatrix},$ $\vec{v_2}=\begin{pmatrix}-1\\a\\b\\2 \end{pmatrix},$ and $\vec{v_1}=\begin{pmatrix}1\\1\\1\\c \end{pmatrix}.$ What are the conditions on the numbers $a,b,c$ so that the three vectors are linearly dependent on $\mathbb{R}^4$? I know that the usual method of solving this is to show that there exists scalars $x_1,x_2,x_3$ not all zero such that
\begin{align}
x_1\vec{v_1}+x_2\vec{v_2}+x_3\vec{v_3}=\vec{0}.
\end{align}
Doing this would naturally lead us to the augmented matrix
\begin{pmatrix}
  1 & -1 & 1 &0\\
  4 & a & 1 &0\\
  1& b & 1 &0\\
  -2 & 2 & c &0\\
\end{pmatrix}
Doing some row reduction would lead us to the matrix
\begin{pmatrix}
  1 & -1 & 1 &0\\
  4 & a & 1 &0\\
  0& b+1 & 0 &0\\
  0 & 0 & c+2 &0\\
\end{pmatrix}
I'm not quite sure how to proceed after this. Do I take cases on when whether $b+1$ or $c+2$ are zero and nonzero? 

Just build a matrix with these vectors as rows and perform row reduce. The vectors will be linearly dependent if at least one row is made of zeros. The idea is that the rank of a matrix is the maximum number of linearly independent rows (or columns), hence, the rows will be linearly dependent if and only if $r(A) < 3$.
$$
\begin{pmatrix} 1 & 4 & 1& -2 \\ -1 & a & b & 2\\ 1 & 1 & 1 & c\end{pmatrix}\to
\begin{pmatrix} 1 & 4 & 1& -2 \\ 0 & a+4 & b+1 & 0\\ 0 & -3 & 0 & c+2\end{pmatrix}\to
\begin{pmatrix} 1 & 4 & 1& -2 \\  0 & -3 & 0 & c+2\\ 0 & a+4 & b+1 & 0\end{pmatrix} \to
$$
$$
\begin{pmatrix} 1 & 4 & 1& -2 \\  0 & -3 & 0 & c+2\\ 0 & 0 & b+1 & \frac{(c+2)(a+4)}{3}\end{pmatrix}
$$
So the vectors are linearly dependent if and only if the last row is filled with zeros, i.e. $b = -1 \wedge (a=-4 \vee c=-2)$.
