Let a, b and c be three unit vectors such that `a+b+c=0`. If `lamda=a*b+b*c+c*a` and `d=axxb+bxxc+cxxa`, then the ordered pair, `(lamda,d)` is equal to :

Options

A. `(3/2,3axxc)`
B. `(-3/2,3cxxb)`
C. `(-3/2, 3axxb)`
D. `(3/2,3bxxc)`

Answer

To solve the problem, we need to find the ordered pair \((\lambda, d)\) given the conditions on the unit vectors \(a\), \(b\), and \(c\) such that \(a + b + c = 0\).

### Step-by-Step Solution

1. **Understanding the Condition**:
We have three unit vectors \(a\), \(b\), and \(c\) such that:
\[ a + b + c = 0 \]
This implies that:
\[ c = - (a + b) \]

2. **Finding \(\lambda\)**:
We know that:
\[ \lambda = a \cdot b + b \cdot c + c \cdot a \]
Substituting \(c = - (a + b)\) into the expression for \(\lambda\):
\[ \lambda = a \cdot b + b \cdot (- (a + b)) + (- (a + b)) \cdot a \]
Simplifying this:
\[ \lambda = a \cdot b - b \cdot a - b \cdot b - a \cdot a - a \cdot b \]
Since \(b \cdot b = 1\) and \(a \cdot a = 1\) (because they are unit vectors):
\[ \lambda = a \cdot b - 1 - 1 - a \cdot b = -2 \]

3. **Finding \(d\)**:
We have:
\[ d = a \times b + b \times c + c \times a \]
Again substituting \(c = - (a + b)\):
\[ d = a \times b + b \times (- (a + b)) + (- (a + b)) \times a \]
Simplifying this:
\[ d = a \times b - b \times a - b \times b - a \times a \]
Since \(b \times b = 0\) and \(a \times a = 0\):
\[ d = a \times b - b \times a = a \times b + a \times b = 2(a \times b) \]

4. **Final Values**:
We have found:
\[ \lambda = -1 \]
\[ d = 2(a \times b) \]
Thus, the ordered pair \((\lambda, d)\) is:
\[ (-1, 2(a \times b)) \]

### Conclusion
The ordered pair \((\lambda, d)\) is:
\[ \left(-1, 2(a \times b)\right) \]