Deduce the condition for the vectors `2hati + 3hatj - 4hatk and 3hati - alpha hatj + bhatk` to be parallel.

Explanation

To determine the condition for the vectors \( \mathbf{A} = 2\hat{i} + 3\hat{j} - 4\hat{k} \) and \( \mathbf{B} = 3\hat{i} - \alpha\hat{j} + b\hat{k} \) to be parallel, we can use the property that two vectors are parallel if their cross product is zero.

### Step-by-Step Solution:

1. **Write the vectors**:
\[ \mathbf{A} = 2\hat{i} + 3\hat{j} - 4\hat{k} \]
\[ \mathbf{B} = 3\hat{i} - \alpha\hat{j} + b\hat{k} \]

2. **Set up the cross product**:
The cross product \( \mathbf{A} \times \mathbf{B} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors \( \mathbf{A} \) and \( \mathbf{B} \):
\[ \mathbf{A} \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -4 \\ 3 & -\alpha & b \end{vmatrix} \]

3. **Calculate the determinant**:
Expanding the determinant:
\[ \mathbf{A} \times \mathbf{B} = \hat{i} \begin{vmatrix} 3 & -4 \\ -\alpha & b \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & -4 \\ 3 & b \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 3 \\ 3 & -\alpha \end{vmatrix} \]

Calculating each of the 2x2 determinants:
- For \( \hat{i} \):
\[ 3b - 4(-\alpha) = 3b + 4\alpha \]
- For \( \hat{j} \):
\[ 2b - (-4)(3) = 2b + 12 \]
- For \( \hat{k} \):
\[ 2(-\alpha) - 3(3) = -2\alpha - 9 \]

4. **Combine the results**:
Thus, the cross product is:
\[ \mathbf{A} \times \mathbf{B} = (3b + 4\alpha)\hat{i} - (2b + 12)\hat{j} + (-2\alpha - 9)\hat{k} \]

5. **Set the cross product equal to zero**:
For the vectors to be parallel, we need:
\[ 3b + 4\alpha = 0 \quad (1) \]
\[ 2b + 12 = 0 \quad (2) \]
\[ -2\alpha - 9 = 0 \quad (3) \]

6. **Solve the equations**:
From equation (2):
\[ 2b + 12 = 0 \implies 2b = -12 \implies b = -6 \]

From equation (3):
\[ -2\alpha - 9 = 0 \implies -2\alpha = 9 \implies \alpha = -\frac{9}{2} \]

7. **Final condition**:
The vectors \( \mathbf{A} \) and \( \mathbf{B} \) are parallel if:
\[ \alpha = -\frac{9}{2}, \quad b = -6 \]