If four points are coplanar then prove that [a b d] + [b c d] + [c a d] = [a b c]. My attempt : 
I assumed the four points A, B, C and D with position vector a, b, c and d to be on the x-y plane. So cross product between any of them is towards the z-axis. 
So I expanded the equation :
$$[\hat a \hat b \hat d] + [\hat b \hat c \hat d] + [\hat c \hat a \hat d] = [\hat a \hat b \hat c]$$
$$\hat a\cdot(\hat b \times \hat c) + \hat b\cdot(\hat c  \times  \hat d) + \hat c\cdot(\hat a  \times  \hat d) = \hat a\cdot(\hat b  \times  \hat c)$$ 
$$0 + 0 + 0 = 0$$
The position vectors are coplanar so their scalar triple product is zero.
But I do not think this method will work in other cases. For example, if I take the points along the x - plane, the position vectors will form a pyramid like structure. So the cross product of any two position vector will not be in a constant direction and the above method fails. 
So can you give me a universal proof which is valid for all the cases.
 A: Partial answer/hint: Actually, your proof is 90% of the way there. For if the vectors all lie in some other plane through the origin, then in a different (orthonormal, oriented) basis, they lie in 
the $z = 0$ plane, and the triple product is invariant under (oriented) orthonormal change of basis. (If this isn't a theorem in your text, you should prove it by showing it's true for rotations in the $xy$, $yz$, and $zx$ planes, since those generate all of $SO(3)$; then you'll have the theorem in your pocket later when you need it.)
The only problem is planes NOT through the origin. For such a plane, there's a point $P$ nearest to the origin, and your argument then applies to $A-P, B-P, C-P, D-P$; you need only show that the terms involving $P$ end up cancelling nicely. 
Hint: Why did I choose the point $P$ to be the one nearest the origin? What's that tell you about $A - P$ relative to $P - O$, where $O$ is the origin? 
A: The vectors $\mathbf{DA}, \mathbf{DB}, \mathbf{DC}$ are coplanar. Hence 
$\mathbf{DA} \times \mathbf{DB} \cdot \mathbf{DC} = 0$ and this gives $$(\mathbf{a}-\mathbf{d}) \times (\mathbf{b}-\mathbf{d})\cdot (\mathbf{c} - \mathbf{d}) = 0$$ Expanding this we obtain the desired expression.
