You know that three points A,B,CA,B,C (two vectors A⃗ BA→B, A⃗ CA→C) form a plane. If you want to show the fourth one DD is on the same plane, you have to show that it forms, with any of the other point already belonging to the plane, a vector belonging to the plane (for instance A⃗ DA→D).
Since the cross product of two vectors is normal to the plane formed by the two vectors (A⃗ B×A⃗ CA→B×A→C is normal to the plane ABCABC), you just have to prove your last vector A⃗ DA→D is normal to this cross product, hence the triple product that should be equal to 00:
A⃗ D⋅(A⃗ B×AC)=0