If the points whose position vectors are `2hati + hatj + hatk , 6hati - hatj + 2 hatk and 14 hati - 5 hatj + p hatk ` are collinear, then p =

Options

A. 2
B. 4
C. 6
D. 8

Answer

To determine the value of \( p \) such that the points with position vectors \( \mathbf{a} = 2\hat{i} + \hat{j} + \hat{k} \), \( \mathbf{b} = 6\hat{i} - \hat{j} + 2\hat{k} \), and \( \mathbf{c} = 14\hat{i} - 5\hat{j} + p\hat{k} \) are collinear, we can use the property that three points are collinear if the determinant of the matrix formed by their position vectors is zero.

### Step-by-step Solution:

1. **Set Up the Determinant**:
We will form a matrix using the coefficients of the position vectors:
\[ \begin{vmatrix} 2 & 1 & 1 \\ 6 & -1 & 2 \\ 14 & -5 & p \end{vmatrix} = 0 \]

2. **Calculate the Determinant**:
We can expand the determinant using the first row:
\[ = 2 \begin{vmatrix} -1 & 2 \\ -5 & p \end{vmatrix} - 1 \begin{vmatrix} 6 & 2 \\ 14 & p \end{vmatrix} + 1 \begin{vmatrix} 6 & -1 \\ 14 & -5 \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
- For the first determinant:
\[ \begin{vmatrix} -1 & 2 \\ -5 & p \end{vmatrix} = (-1)p - (2)(-5) = -p + 10 = 10 - p \]
- For the second determinant:
\[ \begin{vmatrix} 6 & 2 \\ 14 & p \end{vmatrix} = (6)(p) - (2)(14) = 6p - 28 \]
- For the third determinant:
\[ \begin{vmatrix} 6 & -1 \\ 14 & -5 \end{vmatrix} = (6)(-5) - (-1)(14) = -30 + 14 = -16 \]

4. **Substitute Back into the Determinant**:
Substitute the values of the 2x2 determinants back into the determinant:
\[ 2(10 - p) - (6p - 28) - 16 = 0 \]
Simplifying this gives:
\[ 20 - 2p - 6p + 28 - 16 = 0 \]
\[ 20 + 28 - 16 = 2p + 6p \]
\[ 32 = 8p \]

5. **Solve for \( p \)**:
\[ p = \frac{32}{8} = 4 \]

### Final Answer:
Thus, the value of \( p \) is \( \boxed{4} \).