$x-2y+kz=1$ .........(1)
$2x+y+z=2$ .........(2)
$3x-y-kz=3$ ........(3)
 

Hence,

$\left[\begin{array}{ccc}1& -2& k\\ 2& 1& 1\\ 3& -1& -k\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}1\\ 2\\ 3\end{array}\right]$

For infinite solution,

$\left|\begin{array}{ccc}1& -2& k\\ 2& 1& 1\\ 3& -1& -k\end{array}\right|=0$

$\Rightarrow \left(-k+1\right)+2\left(-2k-3\right)+k\left(-2-3\right)=0$

$\Rightarrow k=-\frac{1}{2}$

Put this value in the equation $\left(1\right)$, we get

$2x-4y-z=2....\left(4\right)$

Adding the equation $\left(4\right) \& \left(2\right)$, we get

$4x-3y=4$

Hence, the equation of straight line is $4x-3y-4=0$