Under which one of the following conditions will the two planes `x+y+z=7` and `alphax+betay+gammaz=3` be parallel (but not coincident)?

Options

A. `alpha=beta=gamma=1` only
B. `alpha=beta=gamma=3/7` only
C. `alpha=beta=gamma`
D. None of these

Answer

To determine under which condition the two planes \( x + y + z = 7 \) and \( \alpha x + \beta y + \gamma z = 3 \) are parallel (but not coincident), we need to analyze the equations of the planes.

### Step-by-Step Solution:

1. **Identify the Normal Vectors**:
The normal vector of the first plane \( x + y + z = 7 \) can be derived from its coefficients. The normal vector \( \mathbf{n_1} \) is:
\[ \mathbf{n_1} = (1, 1, 1) \]
The normal vector of the second plane \( \alpha x + \beta y + \gamma z = 3 \) is:
\[ \mathbf{n_2} = (\alpha, \beta, \gamma) \]

2. **Condition for Parallelism**:
Two planes are parallel if their normal vectors are scalar multiples of each other. This means:
\[ \mathbf{n_1} \parallel \mathbf{n_2} \implies (1, 1, 1) \parallel (\alpha, \beta, \gamma) \]
This can be expressed as:
\[ \frac{1}{\alpha} = \frac{1}{\beta} = \frac{1}{\gamma} \]
This implies that:
\[ \alpha = k, \quad \beta = k, \quad \gamma = k \quad \text{for some scalar } k \]

3. **Condition for Non-Coincidence**:
For the planes to be parallel but not coincident, the constant terms must be different. The constant term of the first plane is \( 7 \) and for the second plane, it is \( 3 \). Therefore, we need:
\[ k \neq 7 \quad \text{(since if } k = 7, \text{ the planes would be coincident)} \]

4. **Conclusion**:
The planes \( x + y + z = 7 \) and \( \alpha x + \beta y + \gamma z = 3 \) will be parallel but not coincident if:
\[ \alpha = \beta = \gamma \quad \text{and} \quad \alpha \neq 7 \]

### Final Answer:
The condition for the planes to be parallel but not coincident is that \( \alpha = \beta = \gamma \) and \( \alpha \neq 7 \).