The question.
Let $\xi=(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb R^4$ be a vector with irrational coordinates.
I am interested in finding the minimal value $\mu_\xi$ of
$$\left\vert \det \begin{pmatrix} a_1 & a_2 & 0 & 1 \\ b_1 & b_2 & 1 & 0 \\ c_1 & c_2 & \xi_1 & \xi_3 \\ d_1 & d_2 & \xi_2 & \xi_4 \end{pmatrix} \right\vert,$$
for $a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2\in\mathbb Z$, in terms of the area of the parallelepiped formed by the two vectors $X_i:=(a_i,b_i,c_i,d_i)$, $i=1,2$ (which I assume linearly independent), in $\mathbb R^4$.
Let's call this area $D(X_1,X_2)$. I know we have
$$\begin{align*} D(X_1,X_2)^2 &= \Vert X_1\Vert^2\Vert X_2\Vert^2-(X_1\cdot X_2)^2 \end{align*},$$
but I have no clue on how to proceed from here.
The conjecture.
My hope (which would help the construction of another proof a lot) would be that if we chose the $\xi_i$ properly, we can show that the minimal value verifies
\begin{equation} \mu_\xi\geqslant \frac c{D(X_1,X_2)^2}\qquad\qquad (1) \end{equation}
where $c$ is a constant (it may depends on $\xi$).
Final remarks.
Despite the fact that I strongly believe that $(1)$ is true, any proof that would show that
$$\mu_\xi\geqslant \frac c{D(X_1,X_2)^\gamma}$$
for a $\gamma<4$ would be of great interest.
The answer is wide so we need notations, for any $(X_1,X_2)$ in ${\mathbb{Z}}^2$ distinct, there is $A$ such that $\dfrac{1}{D(X_1,X_2)}\le \dfrac{1}{A}$. $C:=area(X_1,X_2)\ge a$ (parallelogram) also $a\neq 0$ exist. The determinant is the volume of the four column vectors of the given matrix [which we suppose invertible] and equals $\mu_{\xi}=C.\alpha_{\xi}\ge \frac{c_{\xi}}{A}\ge \frac{c_{\xi}}{D(X_1,X_2)} $ for some $\alpha_{\xi}$ where $c_{\xi}= a.A.\alpha_{\xi}$.
Edit $D(X_1,X_2)$ is the volume of $(X_1,X_2, X_1\times X_2)$
Finding the minimum value $a$ explicitly is the same as finding the minimum value of $A$.