Some context.
By rational subspace, I mean a subspace of $\mathbb R^5$ which admits a rational basis. In other words, a basis formed with vectors of $\mathbb Q^5$.
For instance, the vector $v=(0,\pi,3\pi,\pi-\pi^2)$ is in a rational subspace of dimension $2$ since:
$$v=\pi\begin{pmatrix} 0 \\ 1 \\ 3 \\ 1\end{pmatrix}+\pi^2\begin{pmatrix} 0 \\ 0 \\ 0 \\ -1\end{pmatrix}.$$
The question.
Let $A=\mathrm{Span}(Y_1,Y_2,Y_3)$ where
$$Y_1=\begin{pmatrix} 1 \\ 0 \\ 0 \\ \sqrt 6 \\ \sqrt {15} \end{pmatrix},\quad Y_2=\begin{pmatrix} 0 \\ \sqrt{6} \\ 0 \\ -\sqrt {15} \\ \sqrt {10} \end{pmatrix}\quad\text { and }\quad Y_3=\begin{pmatrix} 0 \\ 0 \\ \sqrt{15} \\ \sqrt {10} \\ \sqrt {6} \end{pmatrix}.$$
Does there exist a rational subspace $B$ of $\mathbb R^5$, of dimension $2$, such that $A\cap B\ne\{0\}$?
I believe the answer to be negative.
What I tried.
For $Y\in A$, let's denote by $Y_1,\ldots,Y_5\in\mathbb R$ its coordinates. We can prove the following lemma.
Lemma. The answer to the question is negative, if, and only if,
$$\forall Y \in A,\quad \dim_{\mathbb Q}(\mathrm{Span}_{\mathbb Q}(Y_1,\ldots,Y_5)\ge 3.$$
I tried to investigate what such a rational subspace $B$ would look like. Let's take $B$, a rational subspace of $\mathbb R^5$ of dimension $2$, such that $A\cap B\ne\{0\}$.
Let $Y:=\alpha Y_1+\beta Y_2+\gamma Y_3$ be a vector in $A\setminus\{0\}$.
We have
$$\alpha Y_1+\beta Y_2+\gamma Y_3=\begin{pmatrix} \alpha \\ \beta\sqrt 6 \\ \gamma \sqrt {15} \\ \alpha\sqrt 6-\beta\sqrt {15} +\gamma\sqrt{10} \\ \alpha\sqrt{15}+\beta\sqrt{10}+\gamma\sqrt{6}\end{pmatrix}.$$
If we assume (to try to get somewhere) that
$$\dim_{\mathbb Q}(\mathrm{Span}_{\mathbb Q}(\alpha,\sqrt{6}\beta)=2,$$
then the last three coordinates of $Y$ must be a rational linear combination of the first two, i.e. there exists $x_1,\ldots,x_6\in\mathbb Q$ such that
$$\begin{cases}\gamma \sqrt {15}=x_1\alpha+x_2\beta\sqrt 6 \\ \alpha\sqrt 6-\beta\sqrt {15} +\gamma\sqrt{10}= x_3\alpha+x_4\beta\sqrt 6 \\ \alpha\sqrt{15}+\beta\sqrt{10}+\gamma\sqrt{6}=x_5\alpha+x_6\beta\sqrt 6\end{cases}$$
which can be rewrite
$$MX=0\quad\text{ with } \quad M=\begin{pmatrix} -x_1 & -x_2\sqrt 6 & \sqrt{15} \\ \sqrt{6}-x_3 & -\sqrt{15}-x_4\sqrt 6 & \sqrt{10} \\ \sqrt{15}-x_5 & \sqrt{10}- x_6\sqrt 6 & \sqrt 6 \end{pmatrix}\quad \text{ and }\quad X=\begin{pmatrix} \alpha \\ \beta\\ \gamma\end{pmatrix}.$$
So if $\det(M)\ne 0$, we have won, since $(\alpha,\beta,\gamma)=(0,0,0)$ is the only solution, thus $Y=0$ which is absurd.
Let's compute $\det(M)$ then:
$$\det(M)=A\sqrt{15}+B\sqrt{10}+C\sqrt 6+D,$$
with $A,B,C,D\in\mathbb Q$ depending on the $x_i$. Since
$$\dim_{\mathbb Q}(\mathrm{Span}_{\mathbb Q}(\sqrt 6,\sqrt{10},\sqrt{15}))=3,$$
and $A,B,C,D\in\mathbb Q$, if we assume $\det(M)=0$, we must have
$$A=B=C=D=0.$$
The computations give
$$(\mathscr S)\quad \begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \\ -3x_4x_5+3x_3x_6+3x_1=0 \\ 6x_2-5x_3+15x_4=0 \\ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.\end{cases}$$
The question can now be reformulated as follow:
Does the system $(\mathscr S)$ has rational solutions?
If we try to solve the system, we can end up with this expression:
$\frac{20 \, x_{1} x_{3}^{2} x_{5} + 12 \, x_{1}^{3} - 30 \, x_{1}^{2} x_{3} + 20 \, x_{1}^{2} x_{5} + 30 \, x_{3}^{2} x_{5} - 30 \, x_{1} x_{5}^{2} - 75 \, x_{3} x_{5}^{2} + 186 \, x_{1}^{2} - 225 \, x_{3}^{2} + 120 \, x_{1} x_{5} - 90 \, x_{5}^{2} + 450 \, x_{1} + 1125 \, x_{3} + 180 \, x_{5}}{2 \, x_{1} x_{5} + 5 \, x_{3} x_{5} + 6 \, x_{5}}=0.$
The question is then to understand if this surface in $\mathbb R^3$ has rational points. If you are curious about it, this is what the numerator looks like:
Final remarks.
This question really interests me, but I feel quite stuck about it. May be my whole approach isn't going to help. Any hints, references or piece of solutions would be much appreciated.
If your question boils down to the system,
$$(\mathscr S)\quad \begin{cases} 2x_2x_5-2x_1x_6-6x_6+15=0 \\ -3x_4x_5+3x_3x_6+3x_1=0 \\ 6x_2-5x_3+15x_4=0 \\ 6x_2x_3+6x_1x_4+10x_1-30x_2-15x_5+30=0.\end{cases}$$
then, YES, this system has infinitely many rational solutions. If you let,
$$x_1 = x_4 x_5-x_3 x_6$$ $$x_2 = (5/6)(x_3-3x_4)$$ $$x_5 =\frac{-3(15-6x_6+2x_3x_6^2)}{5x_3-15x_4-6x_4x_6}$$
This satisfies the first 3 equations and the 4th becomes a quadratic in $x_4$,
$$\text{Poly}_1 x_4^2+\text{Poly}_2 x_4+\text{Poly}_3=0$$
You simply solve the linear equation,
$$\text{Poly}_1 = -155 + 25 x_3 - 38 x_6 + 20 x_3 x_6 = 0$$
for $x_6$, thus,
$$x_4 =-\frac{\text{Poly}_3}{\text{Poly}_2}$$
with free parameter $x_3$.