Question:

Adding the first two equations: (2\alpha - \beta + 3\gamma) + (\alpha + \beta - 3\gamma) = 4 - 1 => 3\alpha = 3 => \alpha = 1. Substituting \alpha = 1 into the first two equations: 2 - \beta + 3\gamma = 4 => -\beta + 3\gamma = 2 and 1 + \beta - 3\gamma = -1 => \beta - 3\gamma = -2. These are the same equation. Using the third equation: 5(1) - \beta + 3\gamma = 7 => -\beta + 3\gamma = 2. All equations reduce to \beta - 3\gamma = -2, or \beta = 3\gamma - 2. Since \alpha = 1, the condition \alpha\beta\gamma \leq 0 becomes (1)(3\gamma - 2)(\gamma) \leq 0, which is \gamma(3\gamma - 2) \leq 0. This inequality holds for 0 \leq \gamma \leq 2/3. Since \gamma must be an integer, the only possible value is \gamma = 0. If \gamma = 0, then \beta = 3(0) - 2 = -2. Thus, the only triplet is (1, -2, 0). The number of such triplets is 1.