For u ∉ Q̄ with uū algebraic and the two-orbit space of a, b, α, β, γ, δ with α ≠ β or γ ≠ δ: a spanning 2×3 configuration that no Q̄ + Q̄u + Q̄ū + Q̄w inside it carries

Let $u \in \mathbb{C} \setminus \overline{\mathbb{Q}}$, let $a, b \in \overline{\mathbb{Q}}^\times$ with $a^2 \neq b^2$ and $\alpha, \beta, \gamma, \delta \in \overline{\mathbb{Q}}^\times$, put $g_1 = \alpha/(u - a) + \beta/(u + a)$, $g_2 = \gamma/(u - b) + \delta/(u + b)$, and let $W$ be the $\overline{\mathbb{Q}}$-span of $1, u, u^{-1}, g_1, g_2$. Suppose $\alpha \neq \beta$ or $\gamma \neq \delta$, and that $u\bar u$ is algebraic; put $H_0 = \overline{\mathbb{Q}} + \overline{\mathbb{Q}}u + \overline{\mathbb{Q}}\bar u$. Then the six products of $(1, u^2)$ and $(u^{-1}, g_1, g_2)$ lie in $W$ and span it, and for no $w \in W$ outside $H_0$ does $H_0 + \overline{\mathbb{Q}}w$ carry a configuration.

For $\alpha = \beta$, $\gamma = \delta$ the space is R7's two-pole family, whose configuration is a progression; here the products span the whole space.

Proof. With $x = (1, u^2)$ and $y = (u^{-1}, g_1, g_2)$: $u^2/(u \mp a) = u \pm a + a^2/(u \mp a)$ gives $u^2g_1 = (\alpha+\beta)u + (\alpha-\beta)a + a^2g_1$, and likewise for $g_2$, so the six products lie in $W$; their span contains $1$ because $(\alpha-\beta)a$ or $(\gamma-\delta)b$ is non-zero, hence equals $W$. $x$ is free since $u^2 \notin \overline{\mathbb{Q}}$, and $y$ by DiazModulus.two_orbit_space_excludes_squares_and_reciprocals. Since $\bar u = \rho u^{-1}$, every $H_0 + \overline{\mathbb{Q}}w$ with $w \in W$ lies in $W$, and Theorem B (DiazModulus.circle_point_extension_two_by_three_configuration_iff) with exchange would put $u^2$, $\bar u^2$ (so $u^{-2}$) or some $1/(u - e)$ in $W$, which the same node excludes.

Novelty. Not found in the sources read; it is short.