If the system of equations  $2x+3y-z=0,x+ky-2z=3$ and $2x-y+z=0$  has a non-trivial solution  $\left(x,y,z\right)$then $\frac{x}{y}+\frac{y}{z}+\frac{z}{x}+k$  is equal to

Given system of linear equations 

$\begin{array}{r}2x+3y-z=0\\ x+ky-2z=0\end{array}$

$\text{ And }2x-y+z=0\text{ has a non-trivial solution }(x,y,z)$

$\begin{array}{c}\therefore \mathrm{\Delta }=0\Rightarrow \left|\begin{array}{ccc}2& 3& -1\\ 1& k& -2\\ 2& -1& 1\end{array}\right|=0\\ 2(k-2)-3(1+4)-1(1-2k)=0\\ \Rightarrow 2K-4-15+1+2K=0\\ \Rightarrow 4K=18\Rightarrow K=\frac{9}{2}\end{array}$

So, system of linear equations is 

$\begin{array}{r}2x+3y-z=0-----\left(1\right)\\ 2x+9y-4z=0----\left(2\right)\end{array}$

$\text{ And }2x-y+z=0-----\left(3\right)$

From Eqs. (i) and (ii) , we get 

$6y-3z=0,\frac{y}{z}=\frac{1}{2}$

From Eqs. (i) and (iii), we get 

$4x+2y=0\Rightarrow \frac{x}{y}=-\frac{1}{2}$

so,$\frac{x}{z}=\frac{x}{y}\times \frac{y}{z}=-\frac{1}{4}\Rightarrow \frac{z}{x}=-4$                    $\left[\because \frac{y}{z}=\frac{1}{2}and\frac{x}{y}=-\frac{1}{2}\right]$

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