I have a set of equations to satisfy and I would like to find if they have rational solutions, and if they do, what they are. The equations are: \begin{equation} 1 + \alpha = 3 a \alpha\\ 1 + \beta = 3 b \beta\\ 1 + \gamma = 3 c \gamma\\ 1 + \delta = (a + b + c)\delta\\ \gamma = - \frac{\alpha \sigma^2}{2}\\ 2\beta + 3 \delta - \frac{2 \gamma}{\sigma} = 0\\ \sigma > 0 \end{equation} After some algebra I arrived at this equation: \begin{equation} - \alpha^3 \sigma^3 - \alpha^2\beta\sigma^3 - 11 \alpha^2\beta\sigma^2 + 2 \alpha^2\beta\sigma - 2 \alpha\beta^2\sigma^2 + 4\alpha\beta^2 = 0 \end{equation} I am seeking a solution with all variables $a, b, c, \alpha, \beta, \gamma, \delta, \sigma$ rational, with a preference for $\sigma = 1$.
Update: we can add in another parameter $d$: \begin{equation} 1 + d\alpha = 3 a \alpha\\ 1 + d\beta = 3 b \beta\\ 1 + d\gamma = 3 c \gamma\\ 1 + d\delta = (a + b + c)\delta \end{equation} Again, we require d to be rational.
After noting $\alpha \ne 0$, and writing $t = \alpha/\beta$, your last equation is equivalent to $$ \sigma^3 t^2 + (\sigma^3 + 11 \sigma^2 - 2 \sigma) t + 2 \sigma^2-4 = 0 \tag{1}$$ which has rational solutions if and only if $\sigma \ne 0$ and the discriminant $$ \sigma^2 (\sigma^4 + 14 \sigma^3 + 117 \sigma^2 - 28 \sigma + 4)$$ is a square. Thus we are led to the equation $$ \sigma^4 + 14 \sigma^3 + 117 \sigma^2 - 28 \sigma + 4 - s^2 = 0$$ This is an elliptic curve, which has Weierstrass form $-x^3 + y^2 + 4971 x - 134246$. According to Sage, it has rank $1$, and a generator is $x=-29, y=504$. This corresponds to $\sigma=0, s=2$, but we needed $\sigma \ne 0$. OK, another rational point is $x=-29, y=-504$, which corresponds to $\sigma = 7/8$, $s = 569/64$.
With $\sigma = 7/8$, the solutions of equation (1) are $t = 16/49$ and $t = -79/7$. You can then solve the original equations: e.g. with $t = 16/49$ I find $b$ is arbitrary (but not $1/3$) with $$ \eqalign{\alpha &= \frac{16}{49 (3b-1)}\cr \beta &= \frac{1}{3b-1}\cr \gamma &= - \frac{1}{8(3b-1)}\cr \delta &= - \frac{16}{21(3b-1)}\cr a &= \frac{49 b - 11}{16} \cr c &= 3 - 8 b\cr}$$