Trivial rational solution of a system of hyperplanes

Question

Let us consider a vector space $ V $ over $ \mathbb{Q} $ of dim $6$. We denote all the two dimensional subspace in $ V $ by $ G(2,6) $ (The Grassmanian variety). One can define a map $ p $ from $ G(2,6) $ to $P(\Lambda^{d} V) $ by $ p (U) = u_{1} \wedge u_{2} $, where $ U \in G(2,6) $ and $ u_{1},u_{2} $ be a basis of $ U $. Now we have $ 15 $ Plücker coordinates. We also know that $ G(2,6) $ is the zero set of a system of quadratic Plücker polynomials. Now we define $ 5 $ hyperplanes as a linear combination of Plücker coordinates : $ \alpha_{i} = \sum_{j,k} a_{jk}^{i} p_{jk} $ for all $ 1 \leq i \leq 5 $ , where $ p_{jk} $ are the Plücker coordinates. Also $ a_{jk} \in \mathbb{Q} $.

Now my question is 
 Can there exists 5  hyperplane in this above form such that their intersection with Grassmanian variety has no rational point? 

Answer 1

The answer is positive. Indeed, the space of all codimension 5 linear sections of $G(2,6)$ is parameterized by $G(5,15)$. For each rational point $P \in G(2,6)$ the set of linear sections containing $P$ is a copy of $G(5,14) \subset G(5,15)$. Therefore, the set of all codimension 5 linear sections of $G(2,6)$ containing a rational point is a countable union of proper subvarieties. Taking any point in its complement, one obtains a linear section with no rational points.

EDIT. As @abx explains in the comments this does not answer the OP's question since it is absolutely non clear if the complement contains at least one point defined over $\mathbb{Q}$.