$\begin{vmatrix} 1 & 2 & 1 & -2 & 1 & 4\\ -3 & 5 & 8 & 4 & -3 & 7 \\ 2 & 2 & 2 & -1 & -1 & -1 \\ 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 4 & 2 & -4 & 2 & 8 \\ 3 & 5 & 7 & 11 & 13 & 17 \\ \end{vmatrix} $
I tried to make an upper or under triangle matrix, where the value of the determinant the multiplication of the elements in the diagonal. in the 3.rd row I could be a row which is including only one non-zero element but it also didn't help that much. There is probably one trick what I still can't see. My goal is to find an easy not a mechanical way to calculate the value of this determinant.
Notice the fifth row is twice the first row (viewed as row vectors), by the multi-linearity of the determinant, $\det A=0$.