Find the value of p if `(2 hati + 6 hatj + 27 hatk )xx (hati+ 3 hatj + p hatk) = vec0`

Answer

To find the value of \( p \) such that the cross product

\[ (2 \hat{i} + 6 \hat{j} + 27 \hat{k}) \times (\hat{i} + 3 \hat{j} + p \hat{k}) = \vec{0} \]

we will follow these steps:

### Step 1: Set up the cross product
We will use the determinant form to compute the cross product. The vectors are:

- First vector: \( \vec{A} = 2 \hat{i} + 6 \hat{j} + 27 \hat{k} \)
- Second vector: \( \vec{B} = \hat{i} + 3 \hat{j} + p \hat{k} \)

The cross product can be represented as:

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & 3 & p \end{vmatrix} \]

### Step 2: Calculate the determinant
We will expand the determinant:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} 6 & 27 \\ 3 & p \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 27 \\ 1 & p \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 6 \\ 1 & 3 \end{vmatrix} \]

Calculating each of these 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} 6 & 27 \\ 3 & p \end{vmatrix} = 6p - 81 \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & 27 \\ 1 & p \end{vmatrix} = 2p - 27 \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & 6 \\ 1 & 3 \end{vmatrix} = 6 - 6 = 0 \]

### Step 3: Combine the results
Putting these together, we have:

\[ \vec{A} \times \vec{B} = (6p - 81) \hat{i} - (2p - 27) \hat{j} + 0 \hat{k} \]

### Step 4: Set the cross product equal to the zero vector
Since we want this cross product to equal the zero vector \( \vec{0} \):

\[ (6p - 81) \hat{i} - (2p - 27) \hat{j} + 0 \hat{k} = 0 \]

This gives us two equations:

1. \( 6p - 81 = 0 \)
2. \( -(2p - 27) = 0 \)

### Step 5: Solve the equations
From the first equation:

\[ 6p - 81 = 0 \implies 6p = 81 \implies p = \frac{81}{6} = \frac{27}{2} \]

From the second equation:

\[ -(2p - 27) = 0 \implies 2p - 27 = 0 \implies 2p = 27 \implies p = \frac{27}{2} \]

Both equations give the same result.

### Final Answer
Thus, the value of \( p \) is

\[ \boxed{\frac{27}{2}} \]