Show that the four points (2,0,1), (−1,2,3), (3,2,2) and (3,−6,−3) lie in a plane.

Answer 1

we know that 3 points define a plane. So, pick 3 points and determine the normal direction to their plane.

Now replace any of the points with the 4th point, and determine the new normal.

If the two normals are the same, the 4 points are coplanar.

If we label the direction vectors of the points u,v,w,x, then we need

(w-u)x(w-v) to be parallel to (x-u)x(x-v)

<1,2,1>x<4,0,-1> = <-2,5,-8>
<1,-6,-4>x<4,-8,-6> = <4,-10,16>

the two normals are parallel, so the 4 points are coplanar.

Answer 2

let's take the first 3 points and find the equation of the plane containing them
direction vector #1: (-3,2,2)
direction vector #2: (1,2,1)

normal to these , take the cross-product
= (-2, 5, -8)
or (2, -5, 8) in the opposite direction

equation of plane is
2x - 5y + 8z = c
but (2,0,1) lies on it
4 - 0 + 8 = c = 12

the plane equation using the first three points is
2x - 5y + 8z = 12
test for the 4th point (3,-6,-3)

LS = 2(3) -5(-6) + 8(-3) = 12 = RS

So, yes, all 4 points are coplanar.