question pls help ty

3x + 3y + 9z = 30

-4x + 3y + 5z = 7

7x + 4z = 23

4kx + 4z = 3

If k = 7/4. 

4(7/4)x + 4z = 3

7x + 4z = 3, but 7x + 4z = 23??

That is impossible so it doesn't have any solutions. 

=^._.^=

For a certain value of $k$, the system
$\begin{align*} x + y + 3z &= 10, \\ -4x + 3y + 5z &= 7, \\ kx + z &= 3 \end{align*}$
has no solutions. What is this value of $k$?

$\begin{array}{|rcll|} \hline \begin{vmatrix} 1 & 1 & 3 \\ -4 & 3 & 5 \\ k & 0 & 1 \\ \end{vmatrix} &=& 0 \\\\ k* \begin{vmatrix} 1 & 3 \\ 3 & 5 \\ \end{vmatrix} +1* \begin{vmatrix} 1 & 1 \\ -4 & 3 \\ \end{vmatrix} &=&0 \\\\ k*(1*5-3*3) + 1*(1*3-(-4)*1) &=& 0 \\\\ k*(-4) + 1*7 &=& 0 \\\\ k*(-4) &=& -7 \\\\ k &=& \dfrac{-7}{-4} \\\\ \mathbf{k} &=& \mathbf{ \dfrac{7}{4} } \\ \hline \end{array}$