Let $K/\mathbb{Q}$ be a cyclic cubic field, i.e., $K/\mathbb{Q}$ is a Galois cubic extension. Consider the bilinear form defined on $K^2$ by
$$\displaystyle \mathbf{x} \cdot \mathbf{y} = 2 x_2 y_0 - x_1 y_1 + 2 x_0 y_2,$$
where $\mathbf{x} = (x_0, x_1, x_2)$ and $\mathbf{y} = (y_2, y_1, y_0)$.
Is there a solution to the equation
$$\displaystyle \mathbf{x} \cdot \mathbf{y} = 0$$
with entries of $\mathbf{x}$ being conjugates of the entries of $\mathbf{y}$?
As a ternary quadratic form, your item is isotropic; there are null vectors $\vec{x} = \vec{y}$
Just so you know, and before adjusting for the coefficient 4, the parametrization of integer primitive solutions to $y^2 - zx = 0$ are just $$ x = u^2 \; , \; \; y = uv \; , \; \; z = v^2 $$ for coprime $u,v$
You might be able to do something with that in your cubic field, not sure. As I commented, as a quadratic form your thing resembles $y^2 - 4zx = 0,$ so look at $$ x = u^2 \; , \; \; y = 2uv \; , \; \; z = v^2 $$