Let `a, b and c` be distinct non-negative numbers. If vectos `a hati +a hatj +chatk, hati + hatk and chati +chatj+bhatk` are coplanar, then c is

Options

A. the arithmetic mean of a and b
B. the geometric mean of a and b
C. the harmonic mean of a and b
D. equal to zero

Explanation

To solve the problem, we need to determine the value of \( c \) given that the vectors \( \mathbf{A} = a \hat{i} + a \hat{j} + c \hat{k} \), \( \mathbf{B} = \hat{i} + \hat{k} \), and \( \mathbf{C} = c \hat{i} + c \hat{j} + b \hat{k} \) are coplanar.

### Step-by-Step Solution:

1. **Define the Vectors**:
\[ \mathbf{A} = a \hat{i} + a \hat{j} + c \hat{k} \]
\[ \mathbf{B} = \hat{i} + 0 \hat{j} + \hat{k} \]
\[ \mathbf{C} = c \hat{i} + c \hat{j} + b \hat{k} \]

2. **Set Up the Determinant for Coplanarity**:
The vectors are coplanar if the scalar triple product (determinant) of the vectors is zero:
\[ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0 \]

3. **Calculate the Determinant**:
Expanding the determinant:
\[ = a \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} - a \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} + c \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} \]

Now calculating each of these 2x2 determinants:
- First determinant:
\[ = 0 \cdot b - 1 \cdot c = -c \]
- Second determinant:
\[ = 1 \cdot b - 1 \cdot c = b - c \]
- Third determinant:
\[ = 1 \cdot c - 0 \cdot c = c \]

Putting it all together:
\[ = a(-c) - a(b - c) + c(c) \]
\[ = -ac - ab + ac + c^2 \]
\[ = c^2 - ab \]

4. **Set the Determinant to Zero**:
For coplanarity, we set the determinant to zero:
\[ c^2 - ab = 0 \]
\[ c^2 = ab \]

5. **Solve for \( c \)**:
Taking the square root of both sides:
\[ c = \sqrt{ab} \]

### Conclusion:
Thus, the value of \( c \) is \( \sqrt{ab} \).