Let `bar(PR)=3hati+hatj-2hatk and bar(SQ)=hati-3hatj-4hatk` determine diagonals of a parallelogram PQRS and `bar(PT)=hati+2hatj+3hatk` be another vector. Then the volume of the parallelepiped determined by the vectors `bar(PT),bar(PQ) and bar(PS)` is

Options

A. 5
B. 20
C. 10
D. 30

Explanation

To solve the problem, we need to find the volume of the parallelepiped determined by the vectors \( \bar{PT} \), \( \bar{PQ} \), and \( \bar{PS} \).

### Step-by-Step Solution:

1. **Identify the Given Vectors**:
- \( \bar{PR} = 3\hat{i} + \hat{j} - 2\hat{k} \)
- \( \bar{SQ} = \hat{i} - 3\hat{j} - 4\hat{k} \)
- \( \bar{PT} = \hat{i} + 2\hat{j} + 3\hat{k} \)

2. **Find Vectors \( \bar{PQ} \) and \( \bar{PS} \)**:
- The vector \( \bar{PQ} \) can be found using the formula:
\[ \bar{PQ} = \bar{PR} - \bar{SQ} \]
- Calculate \( \bar{PQ} \):
\[ \bar{PQ} = (3\hat{i} + \hat{j} - 2\hat{k}) - (\hat{i} - 3\hat{j} - 4\hat{k}) = (3 - 1)\hat{i} + (1 + 3)\hat{j} + (-2 + 4)\hat{k} = 2\hat{i} + 4\hat{j} + 2\hat{k} \]

- The vector \( \bar{PS} \) can be found using the formula:
\[ \bar{PS} = \bar{SQ} - \bar{PR} \]
- Calculate \( \bar{PS} \):
\[ \bar{PS} = (\hat{i} - 3\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} - 2\hat{k}) = (1 - 3)\hat{i} + (-3 - 1)\hat{j} + (-4 + 2)\hat{k} = -2\hat{i} - 4\hat{j} - 2\hat{k} \]

3. **Calculate the Cross Product \( \bar{PQ} \times \bar{PS} \)**:
- Set up the determinant:
\[ \bar{PQ} \times \bar{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 2 \\ -2 & -4 & -2 \end{vmatrix} \]
- Calculate the determinant:
\[ = \hat{i} \begin{vmatrix} 4 & 2 \\ -4 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & 2 \\ -2 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & 4 \\ -2 & -4 \end{vmatrix} \]
- This results in:
\[ = \hat{i} (4 \cdot -2 - 2 \cdot -4) - \hat{j} (2 \cdot -2 - 2 \cdot -2) + \hat{k} (2 \cdot -4 - 4 \cdot -2) \]
\[ = \hat{i} (-8 + 8) - \hat{j} (-4 + 4) + \hat{k} (-8 + 8) = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = \vec{0} \]

4. **Find the Volume of the Parallelepiped**:
- The volume \( V \) is given by:
\[ V = |\bar{PT} \cdot (\bar{PQ} \times \bar{PS})| \]
- Since \( \bar{PQ} \times \bar{PS} = \vec{0} \), the volume is:
\[ V = |\bar{PT} \cdot \vec{0}| = 0 \]

### Conclusion:
The volume of the parallelepiped determined by the vectors \( \bar{PT} \), \( \bar{PQ} \), and \( \bar{PS} \) is **0**.