If `cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma`, then `(sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma))` is equal to

Options

A. `1`
B. `-1`
C. `3`
D. `-3`

Explanation

To solve the problem, we start with the given conditions:

1. \( \cos \alpha + \cos \beta + \cos \gamma = 0 \)
2. \( \sin \alpha + \sin \beta + \sin \gamma = 0 \)

We need to find the value of:

\[ \frac{\sin 3\alpha + \sin 3\beta + \sin 3\gamma}{\sin(\alpha + \beta + \gamma)} \]

### Step 1: Use Complex Numbers
We can express the angles in terms of complex exponentials. Let:

\[ A = e^{i\alpha}, \quad B = e^{i\beta}, \quad C = e^{i\gamma} \]

Then, we have:

\[ A + B + C = e^{i\alpha} + e^{i\beta} + e^{i\gamma} \]

### Step 2: Analyze the Sums
From the given conditions, we know:

\[ \cos \alpha + \cos \beta + \cos \gamma = \text{Re}(A + B + C) = 0 \]
\[ \sin \alpha + \sin \beta + \sin \gamma = \text{Im}(A + B + C) = 0 \]

Thus, \( A + B + C = 0 \).

### Step 3: Use the Identity for Cubes
Using the identity for the sum of cubes, we have:

\[ A^3 + B^3 + C^3 - 3ABC = (A + B + C)(A^2 + B^2 + C^2 - AB - AC - BC) \]

Since \( A + B + C = 0 \), we can simplify this to:

\[ A^3 + B^3 + C^3 = 3ABC \]

### Step 4: Express \( \sin 3\alpha + \sin 3\beta + \sin 3\gamma \)
Using the exponential form, we can express:

\[ A^3 = e^{3i\alpha}, \quad B^3 = e^{3i\beta}, \quad C^3 = e^{3i\gamma} \]

Thus:

\[ \sin 3\alpha + \sin 3\beta + \sin 3\gamma = \text{Im}(A^3 + B^3 + C^3) = \text{Im}(3ABC) \]

### Step 5: Calculate \( ABC \)
Since \( A = e^{i\alpha}, B = e^{i\beta}, C = e^{i\gamma} \):

\[ ABC = e^{i(\alpha + \beta + \gamma)} \]

### Step 6: Substitute Back
Now we can substitute back into our expression:

\[ \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3 \sin(\alpha + \beta + \gamma) \]

### Step 7: Final Expression
Now we substitute this into our original expression:

\[ \frac{\sin 3\alpha + \sin 3\beta + \sin 3\gamma}{\sin(\alpha + \beta + \gamma)} = \frac{3 \sin(\alpha + \beta + \gamma)}{\sin(\alpha + \beta + \gamma)} = 3 \]

### Conclusion
Thus, the final answer is:

\[ \boxed{3} \]