Find the volume of a parallelopiped whose edges are represented by the vectors:
`veca =2hati-3hatj-4hatk, vecb=hati+2hatj-2hatk, and vecc=3hati+hatj+2hatk`.

Answer

To find the volume of a parallelepiped defined by the vectors \(\vec{a} = 2\hat{i} - 3\hat{j} - 4\hat{k}\), \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\), and \(\vec{c} = 3\hat{i} + \hat{j} + 2\hat{k}\), we can use the scalar triple product, which can be calculated using the determinant of a 3x3 matrix formed by these vectors.

### Step-by-step Solution:

1. **Set Up the Determinant**:
We will create a matrix using the components of the vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\):
\[ \text{Volume} = |\vec{a} \cdot (\vec{b} \times \vec{c})| = \begin{vmatrix} 2 & -3 & -4 \\ 1 & 2 & -2 \\ 3 & 1 & 2 \end{vmatrix} \]

2. **Calculate the Determinant**:
We can calculate the determinant using the formula:
\[ \text{Determinant} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
where \(a, b, c\) are the elements of the first row, and \(d, e, f, g, h, i\) are the elements of the remaining rows.

For our matrix:
\[ \begin{vmatrix} 2 & -3 & -4 \\ 1 & 2 & -2 \\ 3 & 1 & 2 \end{vmatrix} = 2 \begin{vmatrix} 2 & -2 \\ 1 & 2 \end{vmatrix} - (-3) \begin{vmatrix} 1 & -2 \\ 3 & 2 \end{vmatrix} - 4 \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} \]

3. **Calculate the 2x2 Determinants**:
- For \( \begin{vmatrix} 2 & -2 \\ 1 & 2 \end{vmatrix} = (2)(2) - (-2)(1) = 4 + 2 = 6 \)
- For \( \begin{vmatrix} 1 & -2 \\ 3 & 2 \end{vmatrix} = (1)(2) - (-2)(3) = 2 + 6 = 8 \)
- For \( \begin{vmatrix} 1 & 2 \\ 3 & 1 \end{vmatrix} = (1)(1) - (2)(3) = 1 - 6 = -5 \)

4. **Substitute Back into the Determinant**:
\[ = 2(6) + 3(8) - 4(-5) \]
\[ = 12 + 24 + 20 \]
\[ = 56 \]

5. **Final Volume**:
The volume of the parallelepiped is given by the absolute value of the determinant:
\[ \text{Volume} = |56| = 56 \text{ cubic units} \]

### Final Answer:
The volume of the parallelepiped is \(56\) cubic units.