If `a = i-j+k,b=i-2j-k and c=3i+pj+5k` are coplanar, then p =

Options

A. 6
B. `-6`
C. 2
D. `-2`

Answer

To determine the value of \( p \) such that the vectors \( \mathbf{a} = \mathbf{i} - \mathbf{j} + \mathbf{k} \), \( \mathbf{b} = \mathbf{i} - 2\mathbf{j} - \mathbf{k} \), and \( \mathbf{c} = 3\mathbf{i} + p\mathbf{j} + 5\mathbf{k} \) are coplanar, we will use the condition that the scalar triple product of the vectors must be zero.

### Step-by-Step Solution:

1. **Identify the vectors**:
\[ \mathbf{a} = \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}, \quad \mathbf{b} = \begin{pmatrix} 1 \\ -2 \\ -1 \end{pmatrix}, \quad \mathbf{c} = \begin{pmatrix} 3 \\ p \\ 5 \end{pmatrix} \]

2. **Set up the scalar triple product**:
The scalar triple product can be represented as the determinant of a matrix formed by the vectors:
\[ \begin{vmatrix} \mathbf{a} & \mathbf{b} & \mathbf{c} \end{vmatrix} = 0 \]
This translates to:
\[ \begin{vmatrix} 1 & -1 & 1 \\ 1 & -2 & -1 \\ 3 & p & 5 \end{vmatrix} = 0 \]

3. **Calculate the determinant**:
We can expand this determinant using the first row:
\[ = 1 \cdot \begin{vmatrix} -2 & -1 \\ p & 5 \end{vmatrix} - (-1) \cdot \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & -2 \\ 3 & p \end{vmatrix} \]

4. **Calculate the 2x2 determinants**:
- For the first determinant:
\[ \begin{vmatrix} -2 & -1 \\ p & 5 \end{vmatrix} = (-2)(5) - (-1)(p) = -10 + p = p - 10 \]
- For the second determinant:
\[ \begin{vmatrix} 1 & -1 \\ 3 & 5 \end{vmatrix} = (1)(5) - (-1)(3) = 5 + 3 = 8 \]
- For the third determinant:
\[ \begin{vmatrix} 1 & -2 \\ 3 & p \end{vmatrix} = (1)(p) - (-2)(3) = p + 6 \]

5. **Combine the results**:
Putting it all together, we have:
\[ (p - 10) + 8 + (p + 6) = 0 \]
Simplifying this gives:
\[ 2p + 4 - 10 + 8 = 0 \implies 2p + 2 = 0 \]

6. **Solve for \( p \)**:
\[ 2p = -2 \implies p = -1 \]

### Final Answer:
Thus, the value of \( p \) is:
\[ \boxed{-1} \]