To solve the problem, we need to analyze the two parts of the question step by step.
### Part (i): Proving that \( a + b, b + c, c + a \) are coplanar vectors.
1. **Understanding Coplanarity**:
- Vectors \( a, b, c \) are coplanar if their scalar triple product is zero. The scalar triple product of three vectors \( x, y, z \) is given by \( x \cdot (y \times z) \).
2. **Setting Up the Scalar Triple Product**:
- We need to check the scalar triple product of the vectors \( a + b, b + c, c + a \).
- We compute \( (a + b) \cdot ((b + c) \times (c + a)) \).
3. **Calculating the Cross Product**:
- We first compute the cross product \( (b + c) \times (c + a) \):
\[
(b + c) \times (c + a) = b \times c + b \times a + c \times c + c \times a
\]
- Since \( c \times c = 0 \) (the cross product of any vector with itself is zero), we simplify this to:
\[
(b + c) \times (c + a) = b \times c + b \times a + c \times a
\]
4. **Dot Product with the First Vector**:
- Now we compute the dot product:
\[
(a + b) \cdot (b \times c + b \times a + c \times a)
\]
- This expands to:
\[
(a + b) \cdot (b \times c) + (a + b) \cdot (b \times a) + (a + b) \cdot (c \times a)
\]
5. **Evaluating Each Term**:
- The term \( (a + b) \cdot (b \times a) \) is zero because \( b \times a \) is perpendicular to both \( a \) and \( b \).
- Similarly, \( (a + b) \cdot (c \times a) \) is zero because \( c \times a \) is perpendicular to \( a \).
- Thus, we are left with:
\[
(a + b) \cdot (b \times c)
\]
- Since \( a, b, c \) are coplanar, the scalar triple product \( a \cdot (b \times c) = 0 \).
6. **Conclusion**:
- Therefore, \( (a + b) \cdot ((b + c) \times (c + a)) = 0 \), which implies that the vectors \( a + b, b + c, c + a \) are coplanar.
### Part (ii): Proving that \( a \times b, b \times c, c \times a \) are coplanar vectors.
1. **Setting Up the Scalar Triple Product**:
- We need to check the scalar triple product of the vectors \( a \times b, b \times c, c \times a \).
- We compute \( (a \times b) \cdot ((b \times c) \times (c \times a)) \).
2. **Using the Vector Triple Product Identity**:
- We can use the identity \( x \times (y \times z) = (x \cdot z)y - (x \cdot y)z \).
- Thus, we have:
\[
(b \times c) \times (c \times a) = (b \cdot a)c - (b \cdot c)a
\]
3. **Dot Product with the First Vector**:
- Now we compute:
\[
(a \times b) \cdot ((b \cdot a)c - (b \cdot c)a)
\]
- This expands to:
\[
(b \cdot a)(a \times b) \cdot c - (b \cdot c)(a \times b) \cdot a
\]
4. **Evaluating Each Term**:
- The term \( (b \cdot a)(a \times b) \cdot c \) is zero because \( a \times b \) is perpendicular to both \( a \) and \( b \).
- The term \( (a \times b) \cdot a \) is also zero because \( a \times b \) is perpendicular to \( a \).
5. **Conclusion**:
- Since both terms are zero, we conclude that \( (a \times b) \cdot ((b \times c) \times (c \times a)) = 0 \).
- Therefore, the vectors \( a \times b, b \times c, c \times a \) are coplanar.
### Final Answers:
(i) \( a + b, b + c, c + a \) are coplanar vectors.
(ii) \( a \times b, b \times c, c \times a \) are coplanar vectors.
