What is the value of m if the vectors
`2hat(i)-hat(j)+hat(k), hat(i)+2hat(j)-3hat(k) and 3hat(i)+m hat(j)+5 hat(k)` are coplanar?

Options

A. 16
B. `-4`
C. 4
D. `-8`

Answer

To determine the value of \( m \) for which the vectors \( \mathbf{A} = 2\hat{i} - \hat{j} + \hat{k} \), \( \mathbf{B} = \hat{i} + 2\hat{j} - 3\hat{k} \), and \( \mathbf{C} = 3\hat{i} + m\hat{j} + 5\hat{k} \) are coplanar, we can use the property that three vectors are coplanar if the scalar triple product of the vectors is zero. This can be represented using the determinant of a matrix formed by the components of the vectors.

### Step-by-Step Solution:

1. **Write the vectors in component form:**
- \( \mathbf{A} = (2, -1, 1) \)
- \( \mathbf{B} = (1, 2, -3) \)
- \( \mathbf{C} = (3, m, 5) \)

2. **Set up the determinant:**
The vectors are coplanar if the determinant of the matrix formed by their components is zero:
\[ \begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & m & 5 \end{vmatrix} = 0 \]

3. **Calculate the determinant:**
Expanding the determinant:
\[ = 2 \begin{vmatrix} 2 & -3 \\ m & 5 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} + 1 \begin{vmatrix} 1 & 2 \\ 3 & m \end{vmatrix} \]

Now calculate each of the 2x2 determinants:
- First determinant:
\[ \begin{vmatrix} 2 & -3 \\ m & 5 \end{vmatrix} = (2)(5) - (-3)(m) = 10 + 3m \]
- Second determinant:
\[ \begin{vmatrix} 1 & -3 \\ 3 & 5 \end{vmatrix} = (1)(5) - (-3)(3) = 5 + 9 = 14 \]
- Third determinant:
\[ \begin{vmatrix} 1 & 2 \\ 3 & m \end{vmatrix} = (1)(m) - (2)(3) = m - 6 \]

4. **Combine the results:**
Substitute back into the determinant:
\[ 2(10 + 3m) + 14 + (m - 6) = 0 \]
Simplifying:
\[ 20 + 6m + 14 + m - 6 = 0 \]
Combine like terms:
\[ 20 + 14 - 6 = 28 \quad \text{and} \quad 6m + m = 7m \]
Thus, we have:
\[ 28 + 7m = 0 \]

5. **Solve for \( m \):**
\[ 7m = -28 \quad \Rightarrow \quad m = -4 \]

### Final Answer:
The value of \( m \) for which the vectors are coplanar is \( m = -4 \).