About lineal dependence of sets of vectors in $\mathbb{R}^{4}$ Which of these sets of vectors in $\mathbb{R}^{4}$ are linearly dependent?

\begin{eqnarray}
 &\text{(a)}& (1,2,1,-2), (0,-2,-2,0), (0,2,3,1),(3,0,-3,6)\\
 &\text{(b)}& (4,-4,8,0),(2,2,4,0),(6,0,0,2),(6,3,-3,0)\\
 &\text{(c)}& (4,4,0,0),(0,0,6,6),(-5,0,5,5)
\end{eqnarray}

For this in $(a)$ and $(b)$ I compute the determinants of the matrices and they are different of zero, then, $(a)$ and $(b)$ aren't linear dependence, for $(c)$ I reduce the matrix to an escalar matrix and I find that range is 3, so, I don't find anything set of vector with linear dependence. This is true? or Do I miss something?

Let $S:=\{v_{1},v_{2},\ldots,v_{n}\}\subseteq V$ with $V$ a vector space $n-$dimensional over a field $\mathbb{F}$ so $S$ is linearly independent iff $\det(v_{1},v_{2},\ldots,v_{n})\not=0$. For a) since $\det\begin{bmatrix} 1 & 0 & 0 &3 \\ 2 & -2 & 2 & 0\\ 1 & -2 & 3 & -3\\ -2 & 0 & 1 & 6\end{bmatrix}=-24\not=0$ so linearly independent. For b) since $\det\begin{bmatrix} 4 & 2 & 6 & 6\\ -4 & 2 & 0 & 3\\ 8 & 4 & 0 & -3\\ 0 & 0 & 2 & 0 \end{bmatrix}=480\not=0$ so linearly independent. For c) since $A:=\begin{bmatrix} 4 & 4 & 0  & 0\\ 0 & 0 & 6 & 6\\ 5 & 0 &-5 & 5 \end{bmatrix}\sim \cdots \sim \begin{bmatrix} 4 & 4 & 0 & 0\\ 0 & -5 & -5 & 5\\ 0 & 0 & 6 & 6 \end{bmatrix} \sim \begin{bmatrix} \color{red}{1} & 1 & 0 & 0\\ 0 & \color{red}{1} & 1 & -1\\ 0 & 0 & \color{red}{1} & 1 \end{bmatrix}$ so ${\rm rank}(A)=3$ and since $S:=\{(4,4,0,0),(0,0,6,6),(5,0,-5,5)\}$ has $\#S=3$ then $\#S={\rm rank}(A)$ and hence we have $S$ is linearly independent. Therefore the sets a), b) and c) are linearly independent. In the last part we are using the fact: since $S$ has $n$ vectors so the rank of $A$ to be $n$ in order for $S$ to be linearly independent set.
