\mathbb{F}_{3^{n}}}\zeta_3^{Tr_{n}(-ax+x^d)}\\
    &=\chi_{-a}(D_0)+\zeta_3\chi_{-a}(D_1)+\zeta_3^2\chi_{-a}(D_2)\\
    &=\overline{\chi_a(D_0)}+\zeta_3\overline{\chi_a(D_1)}+\zeta_3^2\overline{\chi_a(D_2)}\\
    &=\overline{\chi_a(D_0)}+\zeta_3\chi_a(D_2)+\zeta_3^2\chi_a(D_1)\\
    &=y_a-2x_a+\zeta_3(x_a+y_a \zeta_3^2)+\zeta_3^2(x_a+y_a \zeta_3)\\
    &=3(y_a-x_a).
    \end{aligned}$$ Since $\widehat{F_1}(a),\widehat{F_1}(-a)\in \{0,\pm3^{(n+s)/2}\}$, $$(x_a,y_a) \in \{(0,0),(0,C),(0,-C),(C,C),(C,0),(C,2C),(-C,-C),(-C,-2C),(-C,0)\}$$ where $C=3^{(n+s-2)/2}$. In the following table values of $\chi_a(D_0),\chi_a(D_1),\chi_a(D_2), W_{F}(a),W_{F}(-a)$ corresponding to each possible $(x_a,y_a)$ tuple is given.

[ | c | c | c | c | c | c | c | c | c | c |]{} & $(0,0)$ & $(0,C)$ & $(0,-C)$ & $(C,C)$ & $(C,0)$ & $(C,2C)$ & $(-C,-C)$ & $(-C,-2C)$ & $(-C,0)$\
$\chi_a(D_0)$ & $0$ & $C$ & $-C$ & $-C$ & $-2C$ & $0$ & $C$ & $0$ & $2C$\
$\chi_a(D_1)$ & $0$ & $C\zeta_3$ & $-C\zeta_3$ & $-C\zeta_3^2$ & $C$ & $C\zeta_3-C\zeta_3^2$ & $C\zeta_3^2$ & $C\zeta_3^2-C\zeta_3$ & $-C$\
$\chi_a(D_2)$ & $0$ & $C\zeta_3^2$ & $-C\zeta_3^2$ & $-C\zeta_3$ & $C$ & $C\zeta_3^2-C\zeta_3$ & $C\zeta_3$ & $C\zeta_3-C\zeta_3^2$ & $-C$\
$\widehat{F_1}(a)$ & $0$ & $0$ & $0$ & $-3C$ & $-3C$ & $-3C$ & $3C$ & $3C$ & $3C$\
$\widehat{F_1}(-a)$ & $0$ & $-3C$ & $3C$ & $-0$ & $-3C$ & $3C$ & $0$ & $-3C$ & $3C$\

The tuples $(C,2C)$ and $(-C,-2C)$ are impossible since $$|\chi_a(D_1)|=\sqrt{x_a^2-x_ay_a+y_a^2}=\sqrt{3C^2}=\sqrt{3^{n+s-1}}$$ which contradicts the fact that $|\chi_a(D_1)|\in \mathbb{Z}$ since $n+s-1$ is odd. Therefore the sets $D_1,D_2$ are PGDS.

Next we will show that the 