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

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.)