If `4bar(i) + (2p)/3 bar(j)+ pbar(k)` is parallel to the vector `bar(i) + 2bar(j) + 3bar(k)`, find p.

Answer

To solve the problem, we need to find the value of \( p \) such that the vector \( 4\mathbf{i} + \frac{2p}{3}\mathbf{j} + p\mathbf{k} \) is parallel to the vector \( \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \).

### Step 1: Set up the condition for parallel vectors
Two vectors \( \mathbf{A} \) and \( \mathbf{B} \) are parallel if their cross product is zero. Therefore, we need to compute the cross product of the vectors:
\[ \mathbf{A} = 4\mathbf{i} + \frac{2p}{3}\mathbf{j} + p\mathbf{k} \]
\[ \mathbf{B} = \mathbf{i} + 2\mathbf{j} + 3\mathbf{k} \]

### Step 2: Compute the cross product
The cross product \( \mathbf{A} \times \mathbf{B} \) can be calculated using the determinant:
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & \frac{2p}{3} & p \\ 1 & 2 & 3 \end{vmatrix} \]

### Step 3: Expand the determinant
Expanding the determinant along the first row, we have:
\[ \mathbf{A} \times \mathbf{B} = \mathbf{i} \begin{vmatrix} \frac{2p}{3} & p \\ 2 & 3 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 4 & p \\ 1 & 3 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 4 & \frac{2p}{3} \\ 1 & 2 \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. For \( \mathbf{i} \):
\[ \begin{vmatrix} \frac{2p}{3} & p \\ 2 & 3 \end{vmatrix} = \frac{2p}{3} \cdot 3 - p \cdot 2 = 2p - 2p = 0 \]

2. For \( \mathbf{j} \):
\[ \begin{vmatrix} 4 & p \\ 1 & 3 \end{vmatrix} = 4 \cdot 3 - p \cdot 1 = 12 - p \]

3. For \( \mathbf{k} \):
\[ \begin{vmatrix} 4 & \frac{2p}{3} \\ 1 & 2 \end{vmatrix} = 4 \cdot 2 - \frac{2p}{3} \cdot 1 = 8 - \frac{2p}{3} \]

Putting it all together:
\[ \mathbf{A} \times \mathbf{B} = 0\mathbf{i} - (12 - p)\mathbf{j} + \left(8 - \frac{2p}{3}\right)\mathbf{k} \]

### Step 4: Set the cross product to zero
For the vectors to be parallel, the cross product must equal zero:
\[ 0\mathbf{i} - (12 - p)\mathbf{j} + \left(8 - \frac{2p}{3}\right)\mathbf{k} = 0 \]

This gives us two equations:
1. From the \( \mathbf{j} \) component:
\[ 12 - p = 0 \implies p = 12 \]

2. From the \( \mathbf{k} \) component:
\[ 8 - \frac{2p}{3} = 0 \implies \frac{2p}{3} = 8 \implies 2p = 24 \implies p = 12 \]

### Final Result
Both equations yield the same solution:
\[ \boxed{12} \]