If the system of linear equations

(1) Since, system of linear equations has non-zero solution

$\therefore \Delta=0$'

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

$\Rightarrow 1\left(9-k^{2}\right)-1\left(3-3 k^{2}\right)+3(1-9)=0$

$\Rightarrow 9-k^{2}-3+3 k^{2}-24=0$

$\Rightarrow 2 k^{2}=18 \Rightarrow k^{2}=9, k=\pm 3$

So, equations are

$x+y+3 z=0$ ...(i)

$x+3 y+9 z=0$...(ii)

$3 x+y+3 z=0$...(iii)

Now, from equation (i)-(ii),

$-2 y-6 z=0 \Rightarrow y=-3 z \Rightarrow \frac{y}{z}=-3$ ...(iv)

Now, from equation (i) - (iii),

$-2 x=0 \Rightarrow x=0$

So, $x+\frac{y}{z}=0-3=-3$