Justify that the vectors are linearly dependent

The space spanned by $5$ vectors is at most (if a,b,c,d,e happen to be linearly independent) $5$ dimensional. You are asking if $6$ vectors in such a space are linearly dependent, but they must be by something called the exchange lemma.

If you select the set $\{v_k \}_{k=1}^{\color{red}{5}}$ you can write

$$ \left(\begin{array}{c}v_1\\ v_2 \\ v_3 \\v_4 \\ v_5 \end{array}\right) = \left(\begin{array}{ccccc} 1 & 1 & 1 & 0 & 0 \\ 2 & 2 & 2 & -1 & 0 \\ 1 & -1 & 0 & 0 & -1 \\ 5 & 6 & -1 & 1 & 1 \\ 1 & 0 & -1 & 0 & 3 \end{array}\right) \left(\begin{array}{c} a \\ b \\ c \\ d \\ e\end{array}\right) $$

Which can be solved for $\{a, \cdots, e \}$ since the array as determinant different from zero (-44). This means it is possible to write $a, \cdots, e$ as a linear combination of $v_1, \cdots, v_5$.

This means that $v_6$ can be written as a linear combination of $\{v_k \}_{k=1}^{\color{red}{5}}$