Find the volume of the parallelepiped determined by the vectors $\vec{a}=\langle 3,5,-1\rangle, \vec{b}=\langle 0,2,4\rangle, \vec{c}=\langle 2,4,1\rangle$. Volume $=$ cubic-units

Step 1 ：We are given three vectors \(\vec{a} = \langle 3,5,-1 \rangle\), \(\vec{b} = \langle 0,2,4 \rangle\), and \(\vec{c} = \langle 2,4,1 \rangle\).

Step 2 ：The volume of a parallelepiped determined by three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) can be found by calculating the scalar triple product of these vectors. The scalar triple product is defined as \(\vec{a} \cdot (\vec{b} \times \vec{c})\), where \(\cdot\) denotes the dot product and \(\times\) denotes the cross product.

Step 3 ：First, we calculate the cross product of \(\vec{b}\) and \(\vec{c}\), which gives us \([-14, 8, -4]\).

Step 4 ：Then, we calculate the dot product of \(\vec{a}\) and the cross product of \(\vec{b}\) and \(\vec{c}\), which gives us the volume of the parallelepiped.

Step 5 ：The volume of the parallelepiped is the absolute value of the scalar triple product, which is 2.

Step 6 ：Final Answer: The volume of the parallelepiped determined by the vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) is \(\boxed{2}\) cubic units.