Let `vec(PR)=3hati+hatj-2hatk and vec(SQ)=hati-3hatj-4hatk` represent the diagonals of the parallelogram PQRS. If `vec(PT)=2hati-hatj+hatk` is another vector, then the volume (in cubic units) of the parallelepiped formed by the vectors `vec(PT),vec(PQ) and vec(PS)` is

Answer

To find the volume of the parallelepiped formed by the vectors \( \vec{PT}, \vec{PQ}, \) and \( \vec{PS} \), we will follow these steps:

### Step 1: Find the vectors \( \vec{PQ} \) and \( \vec{PS} \)

Given the diagonals \( \vec{PR} \) and \( \vec{SQ} \), we can express the vectors \( \vec{PQ} \) and \( \vec{PS} \) in terms of these diagonals.

1. **Using the diagonals**:
- The vector \( \vec{PQ} \) can be expressed as:
\[ \vec{PQ} = \frac{1}{2}(\vec{PR} - \vec{SQ}) \]
- The vector \( \vec{PS} \) can be expressed as:
\[ \vec{PS} = \frac{1}{2}(\vec{SQ} - \vec{PR}) \]

### Step 2: Calculate \( \vec{PQ} \) and \( \vec{PS} \)

1. **Calculate \( \vec{PQ} \)**:
\[ \vec{PQ} = \frac{1}{2}((3\hat{i} + \hat{j} - 2\hat{k}) - (\hat{i} - 3\hat{j} - 4\hat{k})) \]
\[ = \frac{1}{2}((3\hat{i} + \hat{j} - 2\hat{k}) - \hat{i} + 3\hat{j} + 4\hat{k}) \]
\[ = \frac{1}{2}((3 - 1)\hat{i} + (1 + 3)\hat{j} + (-2 + 4)\hat{k}) \]
\[ = \frac{1}{2}(2\hat{i} + 4\hat{j} + 2\hat{k}) = \hat{i} + 2\hat{j} + \hat{k} \]

2. **Calculate \( \vec{PS} \)**:
\[ \vec{PS} = \frac{1}{2}((\hat{i} - 3\hat{j} - 4\hat{k}) - (3\hat{i} + \hat{j} - 2\hat{k})) \]
\[ = \frac{1}{2}((\hat{i} - 3\hat{j} - 4\hat{k}) - 3\hat{i} - \hat{j} + 2\hat{k}) \]
\[ = \frac{1}{2}((-2\hat{i} - 4\hat{j} - 2\hat{k})) \]
\[ = -\hat{i} - 2\hat{j} - \hat{k} \]

### Step 3: Volume of the Parallelepiped

The volume \( V \) of the parallelepiped formed by vectors \( \vec{PT}, \vec{PQ}, \) and \( \vec{PS} \) is given by the scalar triple product:

\[ V = |\vec{PT} \cdot (\vec{PQ} \times \vec{PS})| \]

1. **Calculate \( \vec{PQ} \times \vec{PS} \)**:
\[ \vec{PQ} = \hat{i} + 2\hat{j} + \hat{k} \]
\[ \vec{PS} = -\hat{i} - 2\hat{j} - \hat{k} \]
Using the determinant method for the cross product:
\[ \vec{PQ} \times \vec{PS} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 1 \\ -1 & -2 & -1 \end{vmatrix} \]
\[ = \hat{i}(2 \cdot (-1) - 1 \cdot (-2)) - \hat{j}(1 \cdot (-1) - 1 \cdot (-1)) + \hat{k}(1 \cdot (-2) - 2 \cdot (-1)) \]
\[ = \hat{i}(-2 + 2) - \hat{j}(-1 + 1) + \hat{k}(-2 + 2) \]
\[ = \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = \vec{0} \]

Since the cross product is zero, it indicates that the vectors are coplanar, and thus the volume of the parallelepiped is:

\[ V = 0 \]

### Final Answer
The volume of the parallelepiped formed by the vectors \( \vec{PT}, \vec{PQ}, \) and \( \vec{PS} \) is **0 cubic units**.

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