Context.
The main goal is to find whether or not a subspace of $\mathbb R^5$ of dimension $3$ intersects a rational subspace of dimension $2$. By rational subspace, we mean a subspace of $\mathbb R^5$ which admits a rational basis (i.e. a basis formed with vectors with rational entries). This is what motivates this question.
The question.
Let $Y_1,Y_2,Y_3$ be three vectors of $\mathbb R^5$ such that all the coordinates of the $Y_i$ are in
$$\mathbb Q(\sqrt 2,\sqrt 3,\sqrt 6),$$
i.e. the coordinates of the $Y_i$ are of the form
$$a+b\sqrt 2+c\sqrt 3+d\sqrt 6,\qquad a,b,c,d\in\mathbb Q.$$
Let $X_1,X_2\in\mathbb Q^5$ be two vectors with rational entries such that $(X_1,X_2)$ is free over $\mathbb R$.
Does there exist such $(Y_1,Y_2,Y_3)$ such that for all such $(X_1,X_2)$, the matrix $M\in\mathrm M_5(\mathbb R)$ with columns $Y_1,Y_2,Y_3,X_1,X_2$, i.e.
$$M:=(Y_1\vert Y_2\vert Y_3\vert X_1\vert X_2),$$
satisfies$$\det M\ne 0\quad ?$$
Remarks.
I have tried many choices of vectors $Y_1,Y_2,Y_3$, but it always result in a system of four rational equations that I can not solve. The goal would be to show that the system has no rational solution.
Any ideas or references which would be related to this matter would be of great help.
Best Answer
Okay, so my previous answer was clearly wrong, but now I know why: it is because it is impossible.
The main point is the following: Let $a,b,c,d$ be bilinear skew-symmetric forms from $\mathbb{Q}^5$ to $\mathbb{Q}$. Then, there exists some rational vectors $x,y$ such that $(x,y)$ is free and $a(x,y)=b(x,y)=c(x,y)=d(x,y)=0$.
Proof: A standard reasoning shows that $a$ cannot be nondegenerate, so there exists some nonzero $x$ such that for all $y$ $a(x,y)=0$.
Let $H_b=\{y,b(x,y)=0\}$, $H_c$ and $H_d$ to be the same for $c,d$. These are three hyperplanes of $\mathbb{Q}^5$, so their intersection has dimension at least $2$ so it contains a vector $y$ such that $(x,y)$ is free.
Therefore, $a(x,y)=0$ because $x$ is degenerate for $a$, and $b(x,y)=c(x,y)=d(x,y)=0$ because $y$ is in all hyperplanes.
So how is this related to the OP’s question?
Let $Y_1,Y_2,Y_3$ be vectors in $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})^5$. So $\varphi : (u,v) \in (\mathbb{Q}^5)^2 \longmapsto \det (Y_1,Y_2, Y_3, x,y) \in E=\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{6})$ is bilinear and skew-symmetric.
$E$ is a $\mathbb{Q}$-vector space of dimension $4$, so let $a,b,c,d$ be the coordinates of $\varphi$ in any basis. They are skew-symmetric $\mathbb{Q}$-bilinear forms, so there exists a free pair $(x,y)$ of rational vectors vanishing $a,b,c,d$, thus $\varphi(x,y)=0$.