For what value of m are the vectors `2hat(i)-3hat(j)+4hat(k), hat(i)+3hat(j)-hat(k) and m hat(i)-hat(j)+2 hat(k)` coplanar ?

Options

A. `0`
B. `2/5`
C. `1`
D. `8//5`

Explanation

To determine the value of \( m \) for which the vectors \( \mathbf{A} = 2\hat{i} - 3\hat{j} + 4\hat{k} \), \( \mathbf{B} = \hat{i} + 3\hat{j} - \hat{k} \), and \( \mathbf{C} = m\hat{i} - \hat{j} + 2\hat{k} \) are coplanar, we can use the condition that the scalar triple product of the vectors must be zero.

### Step-by-step Solution:

1. **Write the vectors in component form**:
\[ \mathbf{A} = (2, -3, 4), \quad \mathbf{B} = (1, 3, -1), \quad \mathbf{C} = (m, -1, 2) \]

2. **Set up the determinant for the scalar triple product**:
The scalar triple product can be expressed as the determinant of a matrix formed by the components of the vectors:
\[ \begin{vmatrix} 2 & -3 & 4 \\ 1 & 3 & -1 \\ m & -1 & 2 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ = 2 \begin{vmatrix} 3 & -1 \\ -1 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 1 & -1 \\ m & 2 \end{vmatrix} + 4 \begin{vmatrix} 1 & 3 \\ m & -1 \end{vmatrix} \]

Now, calculate each of the 2x2 determinants:
\[ \begin{vmatrix} 3 & -1 \\ -1 & 2 \end{vmatrix} = (3)(2) - (-1)(-1) = 6 - 1 = 5 \]
\[ \begin{vmatrix} 1 & -1 \\ m & 2 \end{vmatrix} = (1)(2) - (-1)(m) = 2 + m \]
\[ \begin{vmatrix} 1 & 3 \\ m & -1 \end{vmatrix} = (1)(-1) - (3)(m) = -1 - 3m \]

Substitute these back into the determinant:
\[ 2(5) + 3(2 + m) + 4(-1 - 3m) = 0 \]
Simplifying this gives:
\[ 10 + 6 + 3m - 4 - 12m = 0 \]
\[ 12 - 9m = 0 \]

4. **Solve for \( m \)**:
Rearranging gives:
\[ 9m = 12 \implies m = \frac{12}{9} = \frac{4}{3} \]

### Final Answer:
The value of \( m \) for which the vectors are coplanar is:
\[ \boxed{\frac{4}{3}} \]