Find the volume of the parallelopiped with coteminous edges AB, AC and AD, where `A-= (3, 2, 1), B-= (4, 2, 1), C -= (0, 1, 4)` and `D-= (0,0,7)`.

Answer

To find the volume of the parallelepiped with coterminous edges \( \vec{AB}, \vec{AC}, \) and \( \vec{AD} \), we will follow these steps:

### Step 1: Find the vectors \( \vec{AB}, \vec{AC}, \) and \( \vec{AD} \)

1. **Calculate \( \vec{AB} \)**:
\[ \vec{AB} = \vec{B} - \vec{A} = (4, 2, 1) - (3, 2, 1) = (4 - 3, 2 - 2, 1 - 1) = (1, 0, 0) \]

2. **Calculate \( \vec{AC} \)**:
\[ \vec{AC} = \vec{C} - \vec{A} = (0, 1, 4) - (3, 2, 1) = (0 - 3, 1 - 2, 4 - 1) = (-3, -1, 3) \]

3. **Calculate \( \vec{AD} \)**:
\[ \vec{AD} = \vec{D} - \vec{A} = (0, 0, 7) - (3, 2, 1) = (0 - 3, 0 - 2, 7 - 1) = (-3, -2, 6) \]

### Step 2: Set up the determinant for the volume

The volume \( V \) of the parallelepiped can be calculated using the scalar triple product, which is given by the determinant of the matrix formed by the vectors \( \vec{AB}, \vec{AC}, \) and \( \vec{AD} \):

\[ V = |\vec{AB} \cdot (\vec{AC} \times \vec{AD})| = \left| \begin{vmatrix} 1 & 0 & 0 \\ -3 & -1 & 3 \\ -3 & -2 & 6 \end{vmatrix} \right| \]

### Step 3: Calculate the determinant

1. **Calculate the determinant**:
\[ \begin{vmatrix} 1 & 0 & 0 \\ -3 & -1 & 3 \\ -3 & -2 & 6 \end{vmatrix} \]
Expanding the determinant along the first row:
\[ = 1 \cdot \begin{vmatrix} -1 & 3 \\ -2 & 6 \end{vmatrix} - 0 + 0 \]
Now calculate the 2x2 determinant:
\[ = 1 \cdot (-1 \cdot 6 - 3 \cdot -2) = 1 \cdot (-6 + 6) = 1 \cdot 0 = 0 \]

### Step 4: Conclusion

The volume of the parallelepiped is:
\[ V = |0| = 0 \]