1. For the homogeneous system

   $$\begin{cases} x+y+3z=0\\ x+3y+k^2 z=0\\ 3x+y+3z=0 \end{cases}$$

   to have a non-zero solution, the determinant of the coefficient matrix must be zero.

2. The coefficient matrix is

   $$A=\begin{pmatrix} 1&1&3\\ 1&3&k^2\\ 3&1&3 \end{pmatrix}$$

   So,

   $$\det(A)=\begin{vmatrix} 1&1&3\\ 1&3&k^2\\ 3&1&3 \end{vmatrix}$$

3. Expand along the first row:

   $$\det(A)=1\begin{vmatrix}3&k^2\\1&3\end{vmatrix}-1\begin{vmatrix}1&k^2\\3&3\end{vmatrix}+3\begin{vmatrix}1&3\\3&1\end{vmatrix}$$ $$=1(9-k^2)-1(3-3k^2)+3(1-9)$$ $$=9-k^2-3+3k^2-24$$ $$=2k^2-18=2(k^2-9)$$

   For non-zero solution,

   $$\det(A)=0 \implies k^2=9$$

4. Now use $k^2=9$ in the equations:

   $$x+y+3z=0 \quad ...(1)$$ $$x+3y+9z=0 \quad ...(2)$$ $$3x+y+3z=0 \quad ...(3)$$

5. Subtract (1) from (3):

   $$(3x+y+3z)-(x+y+3z)=0$$ $$2x=0 \implies x=0$$

6. Put $x=0$ in (1):

   $$y+3z=0 \implies y=-3z$$

   Hence,

   $$\frac{y}{z}=-3$$

7. Therefore,

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

8. Checking options:

   • A: $9$ ✗
   • B: $3$ ✗
   • C: $-9$ ✗
   • D: $-3$ ✓

So the correct answer is D.