To compute the volume of the parallelepiped induced by the vectors $\vec{v}_1$, $\vec{v}_2$, and $\vec{v}_3$, we need to calculate the absolute value of the scalar triple product of these vectors, which is equivalent to the absolute value of the determinant of the matrix formed by placing these vectors as columns.

The vectors given are: $\vec{v}_1 = \begin{bmatrix} -3 \\ 1 \\ 3 \end{bmatrix}, \quad \vec{v}_2 = \begin{bmatrix} -1 \\ 2 \\ 2 \end{bmatrix}, \quad \vec{v}_3 = \begin{bmatrix} 0 \\ -1 \\ -3 \end{bmatrix}$

Form the matrix $A$ using these vectors as columns: $A = \begin{bmatrix} -3 & -1 & 0 \\ 1 & 2 & -1 \\ 3 & 2 & -3 \end{bmatrix}$

The volume $V$ of the parallelepiped is given by: $V = | \det(A) |$

Let's calculate this determinant:

$\det(A) = -3 \begin{vmatrix} 2 & -1 \\ 2 & -3 \end{vmatrix} - (-1) \begin{vmatrix} 1 & -1 \\ 3 & -3 \end{vmatrix} + 0 \cdot \begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix}$

1. First minor: $\begin{vmatrix} 2 & -1 \\ 2 & -3 \end{vmatrix} = (2)(-3) - (2)(-1) = -6 + 2 = -4$ So, $-3 \times (-4) = 12$.

2. Second minor: $\begin{vmatrix} 1 & -1 \\ 3 & -3 \end{vmatrix} = (1)(-3) - (-1)(3) = -3 + 3 = 0$ So, $-1 \times 0 = 0$.

Adding these results together: $\det(A) = 12 + 0 = 12$

Thus, the volume $V$ is: $V = |12| = 12$

Answer:

The volume of the parallelepiped is $\boxed{12}$.

Would you like further details on this solution, or do you have any questions?

Here are some related questions to explore:

1. How would the process change if one of the vectors was zero?
2. What does a negative determinant imply about the orientation of the vectors?
3. How do you find the volume of a parallelepiped in higher dimensions?
4. How would you interpret the volume if given vectors are linearly dependent?
5. What is the geometric significance of the determinant in this context?

Tip: The scalar triple product of three vectors provides the signed volume of the parallelepiped formed by the vectors. The sign indicates the orientation in space.