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For a certain value of k , the system
\(x + y + 3z = 10 \\ -4x + 2y + 5z = 7 \\ kx + z = 3 \)
has no solutions.

What is this value of k ?

\(\begin{array}{|rcll|} \hline \begin{vmatrix} 1 & 1 & 3 \\ -4 & 2 & 5 \\ k & 0 & 1 \\ \end{vmatrix} &=& 0 \\\\ 1\cdot \begin{vmatrix}2&5\\ 0&1 \\\end{vmatrix} -1\cdot\begin{vmatrix}-4&5\\ k&1 \\\end{vmatrix} +3\cdot\begin{vmatrix}-4&2\\ k&0 \\\end{vmatrix} &=& 0 \\\\ 2-(-4-5k)+3(-2k) &=& 0 \\ 2+4+5k-6k &=& 0 \\ 6-k &=& 0 \\ \mathbf{k} & \mathbf{=}& \mathbf{6} \\ \hline \end{array} \)

x + y + 3z = 10   ⇒   -2x - 2y - 6z  =  -20    (1)
-4x + 2y + 5z = 7    (2)

kx + z = 3

Add   (1)  and (2)   and we get that

-6x - z  =  -13   ⇒    6x + z  = 13

When k =  6....then we have this system

6x + z = 13

6x + z  = 3

But this is impossible.....so.....k = 6