Which one of the following planes is normal to the plane `3x+y+z=5` ?

Options

A. `x+2y+z=6`
B. `x-2y+z=6`
C. `x+2y-z=6`
D. `x-2y-z=6`

Answer

To determine which plane is normal to the plane given by the equation \(3x + y + z = 5\), we will follow these steps:

### Step 1: Identify the normal vector of the given plane
The equation of the plane can be expressed in the standard form \(Ax + By + Cz = D\). Here, \(A = 3\), \(B = 1\), and \(C = 1\). The normal vector \( \mathbf{n_1} \) of the plane is given by the coefficients of \(x\), \(y\), and \(z\).

\[ \mathbf{n_1} = 3\mathbf{i} + 1\mathbf{j} + 1\mathbf{k} \]

### Step 2: Check the normal vectors of the options
We need to check the normal vectors of the options provided to see which one is orthogonal to \( \mathbf{n_1} \). Two vectors are orthogonal if their dot product is zero.

### Step 3: Calculate the dot product for each option
Assuming the options are given as follows:
- Option A: \( \mathbf{n_2} = \mathbf{i} + 2\mathbf{j} + \mathbf{k} \)
- Option B: \( \mathbf{n_2} = \mathbf{i} - 2\mathbf{j} - \mathbf{k} \)
- Option C: \( \mathbf{n_2} = \mathbf{i} + 2\mathbf{j} - \mathbf{k} \)
- Option D: \( \mathbf{n_2} = \mathbf{i} - 2\mathbf{j} - \mathbf{k} \)

#### For Option A:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (3)(1) + (1)(2) + (1)(1) = 3 + 2 + 1 = 6 \quad (\text{not orthogonal}) \]

#### For Option B:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (3)(1) + (1)(-2) + (1)(-1) = 3 - 2 - 1 = 0 \quad (\text{orthogonal}) \]

#### For Option C:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (3)(1) + (1)(2) + (1)(-1) = 3 + 2 - 1 = 4 \quad (\text{not orthogonal}) \]

#### For Option D:
\[ \mathbf{n_1} \cdot \mathbf{n_2} = (3)(1) + (1)(-2) + (1)(-1) = 3 - 2 - 1 = 0 \quad (\text{orthogonal}) \]

### Conclusion
The planes corresponding to options B and D are normal to the plane \(3x + y + z = 5\).