Let's convert these points into vectors by joining each one to the origin (0,0,0), then label the vectors:
a=<2,-1,1>, b=<1,3,-2>, c=<-2,1,-3>, d=<3,2,0>. We can also label the corresponding points as A, B, C, D. The line AB not only joins points A and B, it's also a vector, written as:
AB=a-b=<2-1,-1-3,1+2>=<1,-4,3>.
Similarly we can write: BC=b-c=<3,2,1>.
These three points, A, B and C, define the plane. The normal to the plane is the same normal for the vectors AB and BC. To find this normal we use the vector cross product to generate the normal. AB×BC=n=
| i j k |
| 1 -4 3 | = <-10,8,14>
| 3 2 1 |
If point D is coplanar with A, B and C, then AD (for example) will have the same normal. AD=<-1,-3,1>. (We could have used BD or CD instead of AD.)
The vector dot product of AD with the normal we just found will only be zero if D is in the same plane.
n.AD=0=<-10,8,14>.< -1,-3,1>=10-24+14=0. So the four points are coplanar.
(Just as a check we should carry out the dot product of n.AB, n.BC:
<-10,8,14>.< 1,-4,3>=-10-32+42=0; <-10,8,14>.< 3,2,1>=-30+16+14=0.)