Finding irrational entries such that the determinant will never be zero

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 $\varphi(x,y)=0$.