Let `a,b,c` be distinct non-negative numbers. If the vectors `ahati+ahatj+chatk, hati+hatk` and `chati+chatj+bhatk` lies in a plane then `c` is

Options

A. the AM of a and b
B. the GM of a and b
C. the HM 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 \( \vec{A} = a \hat{i} + a \hat{j} + c \hat{k} \), \( \vec{B} = \hat{i} + \hat{k} \), and \( \vec{C} = c \hat{i} + c \hat{j} + b \hat{k} \) lie in the same plane.

### Step-by-step Solution:

1. **Understanding the Condition for Coplanarity**:
Three vectors \( \vec{A}, \vec{B}, \vec{C} \) are coplanar if the scalar triple product is zero, which can be represented using the determinant of a matrix formed by these vectors.

2. **Forming the Matrix**:
The vectors can be expressed in matrix form as follows:
\[ \begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} \]

3. **Calculating the Determinant**:
We will calculate the determinant of the above matrix. The determinant is given by:
\[ D = a \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} - a \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} + c \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} \]

- The first determinant:
\[ \begin{vmatrix} 0 & 1 \\ c & b \end{vmatrix} = 0 \cdot b - 1 \cdot c = -c \]

- The second determinant:
\[ \begin{vmatrix} 1 & 1 \\ c & b \end{vmatrix} = 1 \cdot b - 1 \cdot c = b - c \]

- The third determinant:
\[ \begin{vmatrix} 1 & 0 \\ c & c \end{vmatrix} = 1 \cdot c - 0 \cdot c = c \]

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

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

5. **Solving for \( c \)**:
This implies:
\[ c^2 = ab \]
Taking the square root (since \( c \) is non-negative):
\[ c = \sqrt{ab} \]

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