tuples $(C,0)$ and $(-C,0)$ are also impossible.

First note that $\widehat{F_1}(0)=0$ since the function $Tr(x^d)$ is balanced. For a non-zero element $a$ the following holds $$\begin{split}
    \widehat{F_1}(a)\widehat{F_1}(-a) =& \chi_a(D_0)\chi_a(D_0)+\zeta_p\chi_a(D_0)\chi_a(D_2)+\zeta_p^2(\chi_a(D_0)\chi_a(D_1))\\
    &+\zeta_p\chi_a(D_0)\chi_a(D_1)+\zeta_p^2\chi_a(D_1)\chi_a(D_2)+(\chi_a(D_1)\chi_a(D_1))\\
    &+\zeta_p^2(\chi_a(D_0)\chi_a(D_2))+(\chi_a(D_2)\chi_a(D_2)+\zeta_p\chi_a(D_1)\chi_a(D_2)\\
    &=\left( \chi_a(D_0)\chi_a(D_0)+ \chi_a(D_0)\chi_a(D_1)+ \chi_a(D_1)\chi_a(D_1)\right) (2-\zeta_p-\zeta_p^2)\\
    &=3\left( \chi_a(D_0)\chi_a(D_0)+ \chi_a(D_0)\chi_a(D_1)+ \chi_a(D_1)\chi_a(D_1)\right)\\
    &=3\left( \chi_a(D_0)\chi_a(D_0)- \chi_a(D_1)\chi_a(D_2)\right)\\
    \end{split}$$ We need the following auxiliary lemma.

Let $S$ be a $k-$subset of an abelian group $G$ of order $v$. Then $$\displaystyle \sum_{i=0}^{v-1}\chi_i(SS^{-1})= \sum_{i=0}^{v-1}\chi_i(ke+\sum_{g\in G-{e}}a_gg)=vk.$$

In the group ring $\mathbb{Z}G$ the product $SS^{-1}=k\cdot e+\sum_{g\in G-{e}}a_g\cdot g$ where $a_g \in \mathbb{Z}$. Then $$\begin{split}
        \displaystyle \sum_{i=0}^{v-1}\chi_i(SS^{-1}) &= \sum_{i=0}^{v-1}\chi_i(ke+\sum_{g\in G-{e}}a_gg)\\
        &= \sum_{i=0}^{v-1}k\cdot \chi_i(e)+\sum_{g\in G-{e}}a_g\cdot  \sum_{i=0}^{v-1}\chi_i(g)\\
        &=vk
        \end{split}$$ This holds since for any $g\in G-{e}$ we have $\sum_{i=0}^{v-1}\chi_i(g)=0$ and $\sum_{i=0}^{v-1}\chi_i(e)=v.$

Using the previous lemma for $S=D_0,D_1$ separately, we obtain $$\begin{split}
    \sum_{a\in \mathbb{F}_{3^n}}\widehat{F_1}(a)\widehat{F_1}(-a)&=3\sum_{a\in \mathbb{F}_{3^n}}\left( \chi_a(D_0)\chi_a(D_0)- \chi_a(D_1)\chi_a(D_2)\right)\\
    &=3\sum_{a\in \mathbb{F}_{3^n}} \chi_a(D_0D_0^{-1})- 3\sum_{a\in \mathbb{F}_{3^n}}\chi_a(D_1D_1^{-1})\\
    &=3\cdot 3^{n}\cdot 3^{n-1}-3\cdot 3^{n}\cdot 3^{n-1}\\
    &=0
    \end{split}$$ On the other hand, using the values from the table, we can also write $$\sum_{a\in \mathbb{F}_{3^n}}\widehat{F_1}(a)\widehat{F_1}(-a)= 9(\Lambda_1 +\Lambda_2)C^2$$ where $\Lambda_1=|\{a \in {{\mathbb F}}_{3^n}^*:(x_a,y_a)=(C,0)\}|, \Lambda_2=|\{a \in {{\mathbb F}}_{3^n}^*:(x_a,y_a)=(-C,0)\}|$. This implies that for any $a\in {{\mathbb F}}_{3^n}^*$, $(x_a,y_a) \ne (C,0), (-C,0)$ and hence $D_0$ is also a PGDS.

By Theorem \[cross-plateaued\] we have PGDS with parameters $(v=p^{2n}, k=p^n, \alpha=p^{n}-p^s, \beta=p^{s+n}+p^{n}-p^s)$. If $p=3$ then by Theorem \[partitioned\] 