If the vectors `2 hat i-3 hat j , hat i+ hat j- hat ka n d3 hat i- hat k` form three concurrent edges of a parallelepiped, then find the volume of the parallelepiped.

Answer

To find the volume of the parallelepiped formed by the vectors \( \mathbf{A} = 2\hat{i} - 3\hat{j} \), \( \mathbf{B} = \hat{i} + \hat{j} - \hat{k} \), and \( \mathbf{C} = 3\hat{i} - \hat{k} \), we will use the scalar triple product, which is given by the determinant of a matrix formed by these vectors.

### Step-by-step Solution:

1. **Identify the vectors**:
\[ \mathbf{A} = 2\hat{i} - 3\hat{j} \\ \mathbf{B} = \hat{i} + \hat{j} - \hat{k} \\ \mathbf{C} = 3\hat{i} - \hat{k} \]

2. **Set up the matrix for the scalar triple product**:
The volume \( V \) of the parallelepiped can be calculated using the determinant of the following matrix:
\[ V = |\mathbf{A} \cdot (\mathbf{B} \times \mathbf{C})| = \begin{vmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{vmatrix} \]

3. **Calculate the determinant**:
We will expand this determinant. Let's expand along the third column:
\[ V = 0 \cdot \text{Cofactor} + (-1) \cdot \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} + (-1) \cdot \begin{vmatrix} 2 & -3 \\ 3 & 0 \end{vmatrix} \]

- The first 2x2 determinant:
\[ \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} = (2 \cdot 1) - (-3 \cdot 1) = 2 + 3 = 5 \]

- The second 2x2 determinant:
\[ \begin{vmatrix} 2 & -3 \\ 3 & 0 \end{vmatrix} = (2 \cdot 0) - (-3 \cdot 3) = 0 + 9 = 9 \]

Thus, substituting back into the determinant:
\[ V = 0 - 5 - 9 = -14 \]

4. **Calculate the volume**:
Since volume cannot be negative, we take the absolute value:
\[ V = | -14 | = 14 \]

### Final Answer:
The volume of the parallelepiped is \( 14 \) cubic units.