If the vectors `alpha hat(i) + alpha hat(j) + gamma hat(k). hat(i) + hat(k) and gamma hat(i) + gamma hat(j) + beta hat(k)` lie on a plane. Where `alpha, beta and gamma` are distinct non-negative numbers, they `gamma` is:

Options

A. Arithmetic mean of `alpha and beta`
B. Geometric mean of `alpha and beta`
C. Harmonic mean of `alpha and beta`
D. None of the above

Answer

To solve the problem, we need to determine the relationship between the distinct non-negative numbers \( \alpha \), \( \beta \), and \( \gamma \) given that the vectors \( \hat{a} = \alpha \hat{i} + \alpha \hat{j} + \gamma \hat{k} \), \( \hat{b} = \hat{i} + \hat{k} \), and \( \hat{c} = \gamma \hat{i} + \gamma \hat{j} + \beta \hat{k} \) are coplanar.

### Step 1: Set up the vectors
We have the following vectors:
- \( \hat{a} = \alpha \hat{i} + \alpha \hat{j} + \gamma \hat{k} \)
- \( \hat{b} = \hat{i} + 0 \hat{j} + \hat{k} \)
- \( \hat{c} = \gamma \hat{i} + \gamma \hat{j} + \beta \hat{k} \)

### Step 2: Write the determinant for coplanarity
The vectors are coplanar if the scalar triple product (determinant) of these vectors is zero. We can express this as:
\[ \begin{vmatrix} \alpha & \alpha & \gamma \\ 1 & 0 & 1 \\ \gamma & \gamma & \beta \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:
\[ \alpha \begin{vmatrix} 0 & 1 \\ \gamma & \beta \end{vmatrix} - \alpha \begin{vmatrix} 1 & 1 \\ \gamma & \beta \end{vmatrix} + \gamma \begin{vmatrix} 1 & 0 \\ \gamma & \gamma \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 0 & 1 \\ \gamma & \beta \end{vmatrix} = 0 \cdot \beta - 1 \cdot \gamma = -\gamma \)
2. \( \begin{vmatrix} 1 & 1 \\ \gamma & \beta \end{vmatrix} = 1 \cdot \beta - 1 \cdot \gamma = \beta - \gamma \)
3. \( \begin{vmatrix} 1 & 0 \\ \gamma & \gamma \end{vmatrix} = 1 \cdot \gamma - 0 \cdot \gamma = \gamma \)

Substituting these back into the determinant:
\[ \alpha (-\gamma) - \alpha (\beta - \gamma) + \gamma^2 = 0 \]
This simplifies to:
\[ -\alpha \gamma - \alpha \beta + \alpha \gamma + \gamma^2 = 0 \]
The \( -\alpha \gamma \) and \( +\alpha \gamma \) cancel out, leading to:
\[ -\alpha \beta + \gamma^2 = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ \gamma^2 = \alpha \beta \]

### Step 5: Conclusion
This implies that \( \gamma \) is the geometric mean of \( \alpha \) and \( \beta \):
\[ \gamma = \sqrt{\alpha \beta} \]

### Final Answer
Thus, the relationship is that \( \gamma \) is the geometric mean of \( \alpha \) and \( \beta \).

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