If the centroid of a triangle with vertices `(alpha, 1, 3), (-2, beta, -5) and (4, 7, gamma)` is the origin then `alpha beta gamma` is equal to …..........

Answer

To solve the problem, we need to find the values of \( \alpha \), \( \beta \), and \( \gamma \) such that the centroid of the triangle formed by the vertices \( ( \alpha, 1, 3) \), \( (-2, \beta, -5) \), and \( (4, 7, \gamma) \) is the origin \( (0, 0, 0) \).

### Step-by-Step Solution:

1. **Understanding the Centroid Formula**:
The coordinates of the centroid \( (G_x, G_y, G_z) \) of a triangle with vertices \( (x_1, y_1, z_1) \), \( (x_2, y_2, z_2) \), and \( (x_3, y_3, z_3) \) are given by:
\[ G_x = \frac{x_1 + x_2 + x_3}{3}, \quad G_y = \frac{y_1 + y_2 + y_3}{3}, \quad G_z = \frac{z_1 + z_2 + z_3}{3} \]

2. **Setting Up the Equations**:
Given the vertices:
- \( A(\alpha, 1, 3) \)
- \( B(-2, \beta, -5) \)
- \( C(4, 7, \gamma) \)

We know the centroid is the origin, so:
\[ G_x = 0, \quad G_y = 0, \quad G_z = 0 \]

This leads to the following equations:
- For \( G_x \):
\[ \frac{\alpha - 2 + 4}{3} = 0 \implies \alpha + 2 = 0 \implies \alpha = -2 \]
- For \( G_y \):
\[ \frac{1 + \beta + 7}{3} = 0 \implies 1 + \beta + 7 = 0 \implies \beta + 8 = 0 \implies \beta = -8 \]
- For \( G_z \):
\[ \frac{3 - 5 + \gamma}{3} = 0 \implies 3 - 5 + \gamma = 0 \implies -2 + \gamma = 0 \implies \gamma = 2 \]

3. **Calculating the Product**:
Now we have:
- \( \alpha = -2 \)
- \( \beta = -8 \)
- \( \gamma = 2 \)

We need to find \( \alpha \cdot \beta \cdot \gamma \):
\[ \alpha \cdot \beta \cdot \gamma = (-2) \cdot (-8) \cdot 2 \]
Calculating this:
\[ = 16 \cdot 2 = 32 \]

### Final Answer:
Thus, the value of \( \alpha \cdot \beta \cdot \gamma \) is \( \boxed{32} \).