A rather plane question

Any three points are coplanar, so first we will determine the plane shared by points $A,B$ and $C$ and then check to see if point $D$ also lies on this plane.

To find the normal to the plane shared by $A,B$ and $C$ we calculate the cross product of $\vec{AB} = (2,0,-2)$ and $\vec{AC} = (1,2,2)$ using the method described in the link to find a normal vector $\vec{n} = (4, - 6, 4)$ .

The plane equation is then $(x - 1, y - 2, z - 3) \circ (4, -6, 4) = 0 \Longrightarrow 2x - 3y + 2z = 2$ .

Then as $2 \times 2017 - 3 \times 2016 + 2 \times 1008 = 2$ we can conclude that $D$ does lie on this plane, and thus that $\boxed{Yes}$ , the four given points are coplanar.

All four points lie on the plane $2x - 3y + 2z = 2$ , so they are coplanar.