The points with position vectors `alpha hati+hatj+hatk, hati-hatj-hatk, hati+2hatj-hatk, hati+hatj+betahatk` are coplanar if

Options

A. `(1-alpha)(1+beta)=0`
B. `(1-alpha)(1-beta)=0`
C. `(1+alpha)(1+beta)=0`
D. `(1+alpha)(1-beta)=0`

Explanation

To determine the condition for the points with position vectors \( \alpha \hat{i} + \hat{j} + \hat{k}, \hat{i} - \hat{j} - \hat{k}, \hat{i} + 2\hat{j} - \hat{k}, \hat{i} + \hat{j} + \beta \hat{k} \) to be coplanar, we can follow these steps:

### Step 1: Identify the Position Vectors
Let:
- \( A = \alpha \hat{i} + \hat{j} + \hat{k} \)
- \( B = \hat{i} - \hat{j} - \hat{k} \)
- \( C = \hat{i} + 2\hat{j} - \hat{k} \)
- \( D = \hat{i} + \hat{j} + \beta \hat{k} \)

### Step 2: Find the Vectors from One Point to Others
We can choose point \( B \) as a reference point and find the vectors \( \overrightarrow{BA}, \overrightarrow{BC}, \overrightarrow{BD} \):
- \( \overrightarrow{BA} = A - B = (\alpha - 1) \hat{i} + (1 + 1) \hat{j} + (1 + 1) \hat{k} = (\alpha - 1) \hat{i} + 2 \hat{j} + 2 \hat{k} \)
- \( \overrightarrow{BC} = C - B = (1 - 1) \hat{i} + (2 + 1) \hat{j} + (-1 + 1) \hat{k} = 0 \hat{i} + 3 \hat{j} + 0 \hat{k} = 0 \hat{i} + 3 \hat{j} \)
- \( \overrightarrow{BD} = D - B = (1 - 1) \hat{i} + (1 + 1) \hat{j} + (\beta + 1) \hat{k} = 0 \hat{i} + 2 \hat{j} + (\beta + 1) \hat{k} \)

### Step 3: Set Up the Determinant for Coplanarity
The points are coplanar if the scalar triple product of the vectors \( \overrightarrow{BA}, \overrightarrow{BC}, \overrightarrow{BD} \) is zero. This can be represented as a determinant:
\[ \begin{vmatrix} \alpha - 1 & 2 & 2 \\ 0 & 3 & 0 \\ 0 & 2 & \beta + 1 \end{vmatrix} = 0 \]

### Step 4: Calculate the Determinant
Calculating the determinant:
\[ = (\alpha - 1) \begin{vmatrix} 3 & 0 \\ 2 & \beta + 1 \end{vmatrix} - 2 \begin{vmatrix} 0 & 0 \\ 0 & \beta + 1 \end{vmatrix} + 2 \begin{vmatrix} 0 & 3 \\ 0 & 2 \end{vmatrix} \]

Calculating the first determinant:
\[ = (\alpha - 1)(3(\beta + 1) - 0) = 3(\alpha - 1)(\beta + 1) \]

The other two determinants are zero, so we have:
\[ 3(\alpha - 1)(\beta + 1) = 0 \]

### Step 5: Solve for Conditions
From the equation \( 3(\alpha - 1)(\beta + 1) = 0 \), we can conclude:
- \( \alpha - 1 = 0 \) or \( \beta + 1 = 0 \)
- Thus, \( \alpha = 1 \) or \( \beta = -1 \)

### Final Condition
The points are coplanar if:
\[ (\alpha - 1)(\beta + 1) = 0 \]