Introduction
When three vectors are coplanar, it means that they lie on the same plane. In this problem, we need to prove that if three vectors a,b, and c have coplanar vectors a b, b c, and c a, then a, b, and c are coplanar.

Proof
To prove that a,b, and c are coplanar, we can use the scalar triple product. The scalar triple product of three vectors a, b, and c is defined as:

a . (b x c)

where x denotes the cross product and . denotes the dot product. The scalar triple product gives the signed volume of the parallelepiped formed by the three vectors.

If a,b, and c are coplanar, then the volume of the parallelepiped formed by them is zero. This means that the scalar triple product is zero.

Now, let's consider the vectors a b, b c, and c a. These vectors lie on the same plane, so their cross product is normal to that plane. Let n be the normal vector to the plane containing these vectors. Then we have:

n . (a b x b c) = 0
n . (b c x c a) = 0
n . (c a x a b) = 0

Using the properties of the cross product, we can simplify these equations as follows:

n . ((a x b) x (b x c)) = 0
n . ((b x c) x (c x a)) = 0
n . ((c x a) x (a x b)) = 0

Expanding the cross products, we get:

n . (a x b) . (b x c) - (n . (a x b)) (n . (b x c)) = 0
n . (b x c) . (c x a) - (n . (b x c)) (n . (c x a)) = 0
n . (c x a) . (a x b) - (n . (c x a)) (n . (a x b)) = 0

We know that n . (a x b), n . (b x c), and n . (c x a) are all non-zero, since n is normal to the plane containing a b, b c, and c a. Therefore, we can divide both sides of each equation by these non-zero terms to get:

(a x b) . (b x c) = (b x c) . (c x a) = (c x a) . (a x b)

Expanding the dot products, we get:

(a . (b x c)) (b . (c x a)) = (b . (c x a)) (c . (a x b)) = (c . (a x b)) (a . (b x c))

We can simplify this as:

a . (b x c) . (b . (c x a)) = b . (c x a) . (c . (a x b)) = c . (a x b) . (a . (b x c))

Using the scalar triple product, we can write this as:

a . (b x c) . (a . (b x c)) = b .