If `cos alpha + 2 cos beta +3 cos gamma = sin alpha + 2 sin beta + 3 sin gamma y = 0`, then the value of `sin 3alpha + 8 sin 3beta + 27 sin 3gamma ` is

Options

A. `sin(a+ b+ gamma)`
B. `3 sin (alpha + beta + gamma)`
C. ` 18 sin(alpha + beta + gamma)`
D. `sin(alpha+ beta + gamma)`

Explanation

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

\[ \cos \alpha + 2 \cos \beta + 3 \cos \gamma = \sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 \]

### Step 1: Represent the equations in complex form
We can express the trigonometric functions in terms of complex exponentials. Let:
- \( A = \cos \alpha + i \sin \alpha \)
- \( B = \cos \beta + i \sin \beta \)
- \( C = \cos \gamma + i \sin \gamma \)

Thus, we can rewrite the equation as:
\[ A + 2B + 3C = 0 \]

### Step 2: Separate real and imaginary parts
From the equation \( A + 2B + 3C = 0 \), we can separate the real and imaginary parts:
- Real part: \( \cos \alpha + 2 \cos \beta + 3 \cos \gamma = 0 \)
- Imaginary part: \( \sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 \)

### Step 3: Apply the identity for cubes
Using the identity for the sum of cubes, we know that:
\[ x^3 + y^3 + z^3 = 3xyz \quad \text{if} \quad x + y + z = 0 \]

Let:
- \( x = A \)
- \( y = 2B \)
- \( z = 3C \)

Then we can write:
\[ A^3 + (2B)^3 + (3C)^3 = 3 \cdot A \cdot (2B) \cdot (3C) \]

### Step 4: Calculate the cubes
Calculating the cubes gives us:
\[ A^3 + 8B^3 + 27C^3 = 18ABC \]

### Step 5: Relate to sine functions
From the earlier steps, we can relate the sine functions:
\[ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma = 18 \sin(\alpha + \beta + \gamma) \]

### Step 6: Conclusion
Since we have established that:
\[ \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma = 18 \sin(\alpha + \beta + \gamma) \]
and given that \( \sin \alpha + 2 \sin \beta + 3 \sin \gamma = 0 \), we can conclude that:

The value of \( \sin 3\alpha + 8 \sin 3\beta + 27 \sin 3\gamma \) is:
\[ \boxed{0} \]