Find the volume of the parallelepiped with vertices $(0,0,0)$, $(3,0,0)$, $(0,5,1)$, $(3,5,1)$, $(2,0,5)$, $(5,0,5)$, $(2,5,6)$, and $(5,5,6)$.
When in $2D$, I usually rely on the Shoelace Formula to solve these types of questions. How do you do this when it's in $3D$?
You can use something called the "three dimensional scalar product". A parallelepiped is uniquely determined by three vectors $u, v, w$, and the volume is given by $|u\cdot(v\times w)|$. In this case, your three vectors are $u = (3, 0, 0), v = (0, 5, 1), w = (2, 0, 5)$.
You'll see this is similar to what you're used to, because for a 2D parallelogram determined by vectors $u, v$, the area is $u\times v$
Edit Just to be sure - are you familiar with the dot product and cross product? Suppose $u = (u_1, u_2, u_3), v = (v_1, v_2, v_3)$. Then $u\cdot v = u_1v_1+u_2v_2+u_3v_3$, and $u\times v = (u_2v_3-u_3v_2, u_3v_1-u_1v_3, u_1v_2-u_2v_1)$