Let $V(J)$ be an irreducible closed subset of $\operatorname{Spec}R[X_1, ..., X_n]$ of dimension $t$. We suppose $R$ is an integral domain with algebraically closed fraction field $K$.
I am wondering about the following. Can we always find $t+1$ linear forms $L_1, ..., L_{t+1}$ with coefficients in $R$ and they are linearly independent over $K$ such that
The dimension of $V(J) \cap V(L_1) \cap ... \cap V(L_j)$ is $t-j$
$V(J) \cap V(L_1) \cap ... \cap V(L_{t+1})$ is the empty set.
Thank you
Edit. The question has been modified based on the commment
No, this is not possible. Clearly there can only be $n$ linearly independent forms in $R[X_1,\cdots,X_n]$ over $K$, so if $t+1>n$ we are out of luck and the claim must be false. Such an example may happen if $R$ is positive-dimensional.
For instance, let $R$ be the ring of algebraic integers (the integral closure of $\Bbb Z$ inside $\overline{\Bbb Q}$). Then $R$ is an integral domain with algebraically closed fraction field and $R$ has dimension one, as integral ring extensions preserve dimension. Now consider $V(0) \subset \Bbb A^1_R$: it is of dimension two, but one cannot find three linearly independent linear forms in $R[X]$.
(An aside: it seems possible to me that this question may be an instance of the XY problem, wherein you're asking about your attempted solution rather than your underlying problem. If this is the case, you may find better help with your underlying issue if you ask a question about your underlying issue and add enough context.)