The system of equations $a_{1}x+b_{1}y+c_{1}z=0, a_{2}x+b_{2}y+c_{2}z=0$ and $a_{3}x+b_{3}y+c_{3}z=0$ has a non-trivial solution, if 

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0.$

Hence, for the given lines $2x+3y-z=0, x+ky-2z=0$ and $2x-y+z=0,$ we have

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

$\Rightarrow 2\left(k-2\right)-3\left(1+4\right)-1\left(-1-2k\right)=0$

$\Rightarrow 2k-4-15+1+2k=0$

$\Rightarrow k=\frac{9}{2}$

Thus, the lines are 

$2x+3y-z=0 ...\left(1\right)$

$x+\frac{9}{2}y-2z=0 ...\left(2\right)$

$2x-y+z=0 ...\left(3\right)$

$\left(1\right)-\left(3\right)\Rightarrow 4y-2z=0$

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

$\frac{y}{z}=\frac{1}{2} ...\left(5\right)$

put $z$ from eq $\left(4\right)$ into $\left(1\right)$

$2x+3y-2y=0$

$2x+y=0$

$\frac{x}{y}=-\frac{1}{2} ...\left(6\right)$

From eq $\left(4\right)$ and $\left(6\right)$

$\frac{z}{x}=-4$

Hence, $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k=-\frac{1}{2}+\frac{1}{2}-4+\frac{9}{2}=\frac{1}{2}.$