Given 3 points, find the area of the triangle/find the volume of the parallelepiped

The points are:

P(-2, 3, 1)

Q(2, -2, 0)

R(4, 1, 0)

I know the norm of the cross product is the area of the parallelogram, so half of this is the area of the triangle. My answer is $\displaystyle \sqrt{497}/2$, but I am not sure if it is right. Can you double check?

Also, I need to find the volume of the parallelepiped formed by OP, OQ, and OR. I assume O is origin, so the vectors are the same as the points right? The scalar triple product is the volume, so the volume is 10. Can you check this too? Don't want to turn in wrong answers...