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.
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
$$\det M\ne 0\quad ?$$
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.