If the system of linear equations
$$x + y + 3z = 0$$
$$x + 3y + k^2z = 0$$
$$3x + y + 3z = 0$$
has a non-zero solution $$(x, y, z)$$ for some $$k \in \mathbb{R}$$, then $$x + \left(\frac{y}{z}\right)$$ is equal to:

We begin by writing the given homogeneous linear system in matrix form. The coefficient matrix is

$$$ \begin{bmatrix} 1 & 1 & 3\\[2pt] 1 & 3 & k^{2}\\[2pt] 3 & 1 & 3 \end{bmatrix}, $$$

and the variable column vector is $$\begin{bmatrix}x\\ y\\ z\end{bmatrix}.$$ For a non-zero (non-trivial) solution to exist, the determinant of the coefficient matrix must vanish. Stating the condition explicitly, we require

$$$\det \begin{bmatrix} 1 & 1 & 3\\ 1 & 3 & k^{2}\\ 3 & 1 & 3 \end{bmatrix}=0.$$$

Now we evaluate this determinant by expanding along the first row. Using the rule

$$$\det \begin{bmatrix} a_{11} & a_{12} & a_{13}\\ a_{21} & a_{22} & a_{23}\\ a_{31} & a_{32} & a_{33} \end{bmatrix} =a_{11}(a_{22}a_{33}-a_{23}a_{32}) -a_{12}(a_{21}a_{33}-a_{23}a_{31}) +a_{13}(a_{21}a_{32}-a_{22}a_{31}), $$$

we substitute the entries:

$$$ \begin{aligned} \det &= 1\bigl(3\cdot3-k^{2}\cdot1\bigr) -1\bigl(1\cdot3-k^{2}\cdot3\bigr) +3\bigl(1\cdot1-3\cdot3\bigr).\\[4pt] \end{aligned} $$$

Calculating each term step by step, we get

$$$ \begin{aligned} 1(9-k^{2}) &= 9-k^{2},\\[2pt] -1(3-3k^{2}) &= -3+3k^{2},\\[2pt] 3(1-9) &= 3(-8)=-24. \end{aligned} $$$

Adding these three expressions,

$$$ (9-k^{2})+(-3+3k^{2})+(-24) =\bigl(9-3-24\bigr)+\bigl(-k^{2}+3k^{2}\bigr) =-18+2k^{2}. $$$

So the determinant equals $$2k^{2}-18.$$ Setting it equal to zero,

$$$2k^{2}-18=0 \;\Longrightarrow\; k^{2}-9=0 \;\Longrightarrow\; k^{2}=9 \;\Longrightarrow\; k=\pm3.$$$

For either value of $$k$$, the second equation becomes

$$x+3y+9z=0.$$

We now solve the three equations with $$k^{2}=9$$:

$$$ \begin{cases} x+y+3z=0,\\[2pt] x+3y+9z=0,\\[2pt] 3x+y+3z=0. \end{cases} $$$

Subtracting the first equation from the second gives

$$$ (x+3y+9z)-(x+y+3z)=0 \;\Longrightarrow\; 2y+6z=0 \;\Longrightarrow\; y+3z=0 \;\Longrightarrow\; y=-3z. $$$

Substituting $$y=-3z$$ into the first equation, we obtain

$$ x+(-3z)+3z=0 \;\Longrightarrow\; x=0. $$

Finally, substituting $$x=0$$ and $$y=-3z$$ into the third equation confirms consistency:

$$ 3(0)+(-3z)+3z=0. $$

Thus a non-zero solution is characterized by

$$ x=0,\qquad y=-3z,\qquad z\neq0. $$

We now compute the required expression:

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

Hence, the correct answer is Option A.