If the system of linear equations
$$x - 2y + kz = 1$$
$$2x + y + z = 2$$
$$3x - y - kz = 3$$
has a solution $$(x, y, z), z \neq 0$$, then $$(x, y)$$ lies on the straight line whose equation is:

First, we write the three given equations clearly:

$$\begin{aligned} x - 2y + kz &= 1 \quad\quad (1)\\[4pt] 2x + y + z &= 2 \quad\quad (2)\\[4pt] 3x - y - kz &= 3 \quad\quad (3) \end{aligned}$$

Because the question asks for the locus of $$(x,y)$$ when a solution with $$z\neq 0$$ exists, we shall eliminate both $$k$$ and $$z$$ from the system. That elimination will leave us with a relation involving only $$x$$ and $$y$$, which will be the required straight-line equation.

We begin by isolating the product $$kz$$ from the first and the third equations. From Equation (1), moving all other terms to the right:

$$kz = 1 - x + 2y \quad\quad (4)$$

Next, from Equation (3) we first move $$3x - y$$ to the right, remembering the sign in front of $$kz$$:

$$ -\,kz = 3 - 3x + y $$

Multiplying both sides by $$-1$$ gives

$$ kz = 3x - y - 3 \quad\quad (5)$$

Now, because both expressions (4) and (5) equal the same quantity $$kz$$, we set them equal to one another:

$$ 1 - x + 2y = 3x - y - 3 $$

We gather like terms on one side. First, move every term to the left:

$$ 1 - x + 2y - 3x + y + 3 = 0 $$

Combining individual groups:

$$ (1+3) + (-x-3x) + (2y + y) = 0 $$

$$ 4 - 4x + 3y = 0 $$

Finally, multiplying the entire equation by $$-1$$ (so that the coefficient of $$x$$ is positive) yields

$$ 4x - 3y - 4 = 0 $$

This linear relation between $$x$$ and $$y$$ must be satisfied by every triplet $$(x,y,z)$$ that solves the original system. Notice that we have not imposed any extra restriction on $$z$$ except that it must be non-zero; the relation we obtained is therefore the complete locus in the $$xy$$-plane.

Equation $$4x - 3y - 4 = 0$$ matches Option A in the list provided.

Hence, the correct answer is Option A.