, 2020

Since the equation is symmetric with respect to $\alpha, \beta, \gamma$ we suppose that $\alpha \ge \beta \ge \gamma$. Then we have:
$$\alpha \le \alpha + \beta + \gamma \le 3\alpha \Leftrightarrow \alpha \le v\alpha\beta\gamma \le 3\alpha \Rightarrow 1 \le v\beta\gamma \le 3.$$

We distinguish the following cases:
• $v > 3$. Then $v\beta\gamma > 3$, absurd.
• $v = 3$. Then $1 \le 3\beta\gamma \le 3 \Rightarrow \beta\gamma = 1 \Rightarrow \beta = \gamma = 1$, and hence
$$\alpha + 2 = 3\alpha \Leftrightarrow \alpha = 1.$$
Therefore we get the solution: $(\alpha, \beta, \gamma) = (1,1,1)$.
• $v = 2$. Then $1 \le 2\beta\gamma \le 3 \Rightarrow \beta\gamma = 1 \Rightarrow \beta = \gamma = 1$, and
$$\alpha + 2 = 2\alpha \Leftrightarrow \alpha = 2.$$
Therefore we get the solution $(\alpha, \beta, \gamma) = (2,1,1)$ and using symmetry we obtain the solutions $(\alpha, \beta, \gamma) = (1,2,1)$ and $(\alpha, \beta, \gamma) = (1,1,2)$.
• $v = 1$. Then $1 \le \beta\gamma \le 3 \Rightarrow \beta\gamma \in \{1,2,3\}$.
If $\beta\gamma = 1$, then $\beta = \gamma = 1$ and $\alpha + 2 = 1$, impossible.
If $\beta\gamma = 2$, then $\beta = 2, \gamma = 1$ and $\alpha + 3 = 2\alpha \Leftrightarrow \alpha = 3$.
Therefore $(\alpha, \beta, \gamma) = (3,2,1)$ and by symmetry
$$\begin{align*} (\alpha, \beta, \gamma) &= (2,1,3), \\ (\alpha, \beta, \gamma) &= (3,2,1), \\ (\alpha, \beta, \gamma) &= (3,1,2), \\ (\alpha, \beta, \gamma) &= (1,2,3), \\ (\alpha, \beta, \gamma) &= (2,3,1). \end{align*}$$
If $\beta\gamma = 3$, then $\beta = 3, \gamma = 1$ and $\alpha + 4 = 3\alpha \Leftrightarrow \alpha = 2$. (rejected, $\alpha < \beta$).
Hence $v \notin \{1,2,3\}$.