Let $X$ be the Thomas plank as defined in the body of the question. Let $\mathcal{Z}(X)$ be the set of zero sets of $X$ i.e. sets of zeros of continuous maps $f\in C(X)$.
Lemma. If $f\in C(X)$ and $Z(f) = \{x\in X : f(x) = 0\}$, then $Z(f)\cap L_0$ is countable or co-countable in $L_0$.
Proof: According to book "Counterexamples in topology" by Steen and Seebach, each $f\in C(X)$ is constant on $L_n$ except for a countable set. For the sake of completion let me prove that as well. Let $A_{m, n} = \{(x, 1/n)\in L_n: |f(x, 1/n)-f(0, 1/n)|\leq 1/m\}$ for $n>0$. The sets $A_{m, n}$ are co-finite in $L_n$, thus $A_n = \bigcap_{m\geq 1}A_{m, n}$ is co-countable. So $f$ is constant on $A_n$. Let $A_0 = \{(x, 0) :\forall_n\ (x, 1/n)\in A_n\}$, then $f$ is constant on $A_0$ and $A_0$ is co-countable. Thus, $f(x) = c\in\mathbb{R}$ for $x\in L_0\setminus A$ where $A\subseteq L_0$ is countable. If $c = 0$, then $L_0\setminus A\subseteq Z(f)$ so that $Z(f)$ is co-countable in $L_0$. If $c\neq 0$, then $Z(f)\cap L_0\subseteq A$ is countable. $\square$
Proposition. Let $\mathcal{U} = \{Z\in \mathcal{Z}(X) : Z\cap L_0\text{ is co-countable in }L_0\}$. Then $\mathcal{U}$ is a free real $z$-ultrafilter on $X$.
Proof: Clearly $X\in\mathcal{U}$ and $\emptyset\notin\mathcal{U}$. If $Z_n\in \mathcal{U}$, then $Z = \bigcap_{n\geq 1} Z_n$ is a zero set, and clearly $Z\cap L_0$ is co-countable as a countable intersection of co-countable sets, so $Z\in\mathcal{U}$. Thus $\mathcal{U}$ is closed under countable intersections. If $Z_0\subseteq Z_1$ are zero sets with $Z_0\in\mathcal{U}$, then $Z_1\cap L_0\supseteq Z_0\cap L_0$ is co-countable, so $Z_1\in\mathcal{U}$. This shows that $\mathcal{U}$ is a $z$-filter closed under countable intersections. To prove that $\mathcal{U}$ is a $z$-ultrafilter, let $Z\notin \mathcal{U}$, so that $Z\cap L_0$ is countable. If $Z\cap L_0$ is finite, then $W = Z\cap L_0\cup \{(x, 1/i) : (x, 0)\in Z\cap L_0, i > 0\}$ is clopen, so since $L_0$ is a zero set (the map $f(x, y) = y$ is continuous), we have that $W^c\cap L_0 = L_0\setminus Z$ is a co-countable zero set disjoint from $Z$. If $Z\cap L_0$ is infinite let $(x_n)_{n\in\mathbb{N}}$ be a sequence such that $\mathbb{N}\ni n \mapsto (x_n, 0)\in Z\cap L_0$ is a bijection. Let $f(x_n, 1/i) = 2^{-n/i}$, $f(x_n, 0) = 1$, $f(x, y) = 0$ otherwise. Then $f \in C(X)$ and $Z(f)\cap L_0 = L_0\setminus Z$ is a co-countable zero set disjoint from $Z$. So $\mathcal{U}$ is a $z$-ultrafilter. Finally, since all points in $L_0$ being compact $G_\delta$ sets are zero sets, and since all of them are countable, none of them belong to $\mathcal{U}$. Of course, since $L_0\in\mathcal{U}$, no other points of $X$ are in all members of $\mathcal{U}$ either. So $\mathcal{U}$ is a free $z$-ultrafilter. $\square$
Corollary. Thomas plank is not realcompact.
Proof: The $z$-ultrafilter $\mathcal{U}$ of previous proposition shows that there are real $z$-ultrafilters on $X$ that aren't fixed, thus $X$ is not realcompact. $\square$
