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

Given system of linear equations 

$\begin{array}{r}x-2y+kz=1----\left(1\right)\\ 2x+y+z=2----\left(2\right)\end{array}$

$\text{ And }3x-y-kz=3----\left(3\right)$

$\text{ Has a solution }(x,y,z),z\ne 0$

On adding Eqs. (i) and (iii), we get 

$\begin{array}{l}x+2y+kz+3x-y-kz=1+3\\ 4x-3y=4\\ \Rightarrow 4x-3y-4=0\end{array}$

This is the required equation of the straight line in which point $\left(x,y\right)$ lies .