If the vectors `4hati+11hatj+mhatk,7hati+2hatj+6hatk and hati+5hatj+4hatk` are coplanar, then m is equal to

Options

A. 38
B. 0
C. 10
D. `-10`

Explanation

To determine the value of \( m \) such that the vectors \( \mathbf{a} = 4\hat{i} + 11\hat{j} + m\hat{k} \), \( \mathbf{b} = 7\hat{i} + 2\hat{j} + 6\hat{k} \), and \( \mathbf{c} = \hat{i} + 5\hat{j} + 4\hat{k} \) are coplanar, we can use the condition that the scalar triple product (or box product) of the vectors must be equal to zero.

### Step 1: Write the vectors
Let:
\[ \mathbf{a} = 4\hat{i} + 11\hat{j} + m\hat{k} \]
\[ \mathbf{b} = 7\hat{i} + 2\hat{j} + 6\hat{k} \]
\[ \mathbf{c} = \hat{i} + 5\hat{j} + 4\hat{k} \]

### Step 2: Set up the determinant
The condition for coplanarity can be expressed as:
\[ \begin{vmatrix} 4 & 11 & m \\ 7 & 2 & 6 \\ 1 & 5 & 4 \end{vmatrix} = 0 \]

### Step 3: Calculate the determinant
Calculating the determinant, we have:
\[ = 4 \begin{vmatrix} 2 & 6 \\ 5 & 4 \end{vmatrix} - 11 \begin{vmatrix} 7 & 6 \\ 1 & 4 \end{vmatrix} + m \begin{vmatrix} 7 & 2 \\ 1 & 5 \end{vmatrix} \]

Calculating the 2x2 determinants:
1. \( \begin{vmatrix} 2 & 6 \\ 5 & 4 \end{vmatrix} = (2 \cdot 4) - (6 \cdot 5) = 8 - 30 = -22 \)
2. \( \begin{vmatrix} 7 & 6 \\ 1 & 4 \end{vmatrix} = (7 \cdot 4) - (6 \cdot 1) = 28 - 6 = 22 \)
3. \( \begin{vmatrix} 7 & 2 \\ 1 & 5 \end{vmatrix} = (7 \cdot 5) - (2 \cdot 1) = 35 - 2 = 33 \)

Substituting these values back into the determinant:
\[ = 4(-22) - 11(22) + m(33) \]
\[ = -88 - 242 + 33m \]
\[ = 33m - 330 \]

### Step 4: Set the determinant to zero
Setting the determinant equal to zero gives:
\[ 33m - 330 = 0 \]

### Step 5: Solve for \( m \)
\[ 33m = 330 \]
\[ m = \frac{330}{33} = 10 \]

Thus, the value of \( m \) is \( 10 \).

### Summary
The value of \( m \) such that the vectors are coplanar is:
\[ \boxed{10} \]