If the system of linear equations $x-2y+kz=1,2x+y+z=2,3x-y-kz=3$ has a solution $\left(x,y,z\right)$, $z\ne 0$  then $\left(x,y\right)$ lies on the straight line whose equation is ____

$\Delta =\left|\begin{array}{l}1-2K\\ 211\\ 3-1K\end{array}\right|=-5\left(2K+1\right);{\Delta }_1=\left|\begin{array}{l}1-2K\\ 211\\ 3-1K\end{array}\right|=-5\left(2K+1\right)$

${\Delta }_2=\left|\begin{array}{l}11K\\ 221\\ 33-K\end{array}\right|=0;{\Delta }_3=\left|\begin{array}{l}1-21\\ 212\\ 3-13\end{array}\right|=0$

$\because Z\ne 0\Rightarrow \Delta =0\Rightarrow K=-\frac{1}{2}$

 $\therefore$system of equation has infinite many solutions.

Let $Z=\lambda \ne 0$ then 

$x=\frac{10-3\lambda }{10}\&y=\frac{-2\lambda }{5}$

 $\therefore$$\left(x,y\right)$ must lie on line  $4x-3y-4=0$