How to find a rational $\mathbb{F}_{\!q}$-curve on a quite classical Calabi–Yau threefold?

Take a finite field $$\mathbb{F}_{\!q}$$ such that $$q \equiv 1 \pmod 3$$, i.e., $$\omega \mathrel{:=} \sqrt[3]{1} \in \mathbb{F}_{\!q}$$, $$\omega \neq 1$$. Also, for $$i \in \{0,1,2\}$$ consider the elliptic curves $$E_i\!: y^2_i = b^ix_i^3 - b$$, where $$b \in \mathbb{F}_{\!q}^* \setminus (\mathbb{F}_{\!q}^*)^3$$. There is on $$E_i$$ the order $$3$$ automorphism $$[\omega]\!: (x_i,y_i) \mapsto (\omega x_i, y_i)$$.

Look at the quotient $$T \mathrel{:=} (E_0 \!\times\! E_1 \!\times\! E_2)/[\omega]^{\times 3}$$, which is a Calabi–Yau threefold according to Oguiso and Truong - Explicit examples of rational and Calabi–Yau threefolds with primitive automorphisms of positive entropy. It is easily seen that it has the affine model $$T\!: \begin{cases} y_1^2 + b = b(y_0^2 + b)t_1^3, \\ y_2^2 + b = b^2(y_0^2 + b)t_2^3 \end{cases} \quad \subset \quad \mathbb{A}^{\!5}_{(y_0,y_1,y_2,t_1,t_2)},$$ where $$t_1 \mathrel{:=} x_1/x_0$$, $$t_2 \mathrel{:=} x_2/x_0$$.

Although $$T$$ is a quite classical quotient, I cannot find a rational $$\mathbb{F}_{\!q}$$-curve on it. In my opinion, this is a sufficiently interesting algebraic geometry task. Can you help me please? I can explain the origin of this task if it is necessary.

Further intersecting $$T$$, which is $$3$$-dimensional in its affine model inside $$\Bbb A^5$$, with a generic variety of codimension two should lead to a curve. (It may be that i do not catch the point of the question. So i am inserting also an example.)
For instance, after multiplying the equations of $$E_0, E_1,E_2$$ by $$1,b^2,b^4$$ we obtain isomorphic curves given by \begin{aligned} Y_0^2 + b^1&= X_0^3\ ,\\ Y_1^2 + b^3&= X_1^3\ ,\\ Y_2^2 + b^5&= X_2^3\ . \end{aligned} Each change of coordinates is linear, so the action of $$[\omega]$$ translates also as a multiplication with $$\omega$$ on the $$X_i$$-components, $$i$$ being $$0,1,2$$. Then the model of the cartesian product of the three curves modulo $$(X_0,X_1,X_2,Y_0,Y_1,Y_2)\sim (\omega X_0,\omega X_1,\omega X_2,Y_0,Y_1,Y_2)$$ is in a similar manner given by the equations: \begin{aligned} \frac{Y_1^2-b^3}{Y_0^2-b} & = \left(\frac {X_1}{X_0}\right)^3=: u_1^3 \ ,\\ \frac{Y_2^2-b^5}{Y_0^2-b} & = \left(\frac {X_2}{X_0}\right)^3=: u_2^3 \ . \end{aligned} (Omit the $$X$$-occurrences and consider the equations in $$\Bbb A^5_{(u_1,u_2;Y_0,Y_1,Y_2)}$$.)
Now we can intersect with the codimension two variety given (in an affine model) by $$Y_1=Y_0^3\ ,\ Y_2=Y_0^5\ .$$ We obtain a curve (of higher degree) parametrized by $$Y_0$$ as follows: \left\{ \begin{aligned} Y_1 &= Y_0^3\ ,\\ Y_2 &= Y_0^5\ ,\\ u_1^3 &= Y_0^4 + bY_0^2+b^2\ ,\\ u_2^5 &= Y_0^8 + bY_0^6 + b^2Y_0^4 + b^3Y_0 + b^4\ . \end{aligned} \right. (Depending on the direction of research this may be useful or not.)
• Well, "rational" is indeed ambiguous here, but I guess the OP means of genus $0$.
• Yes, I mean an $\mathbb{F}_{\!q}$-curve of geometric genus $0$. Jul 28 '20 at 14:14