`a`, `b`,`c` are three complex numbers on the unit circle `|z|=1`, such that `abc=a+b+c`. Then `|ab+bc+ca|` is equal to

Options

A. `3`
B. `6`
C. `1`
D. `2`

Explanation

To solve the problem, we need to find the value of \( |ab + bc + ca| \) given that \( a, b, c \) are complex numbers on the unit circle and satisfy the equation \( abc = a + b + c \).

### Step-by-Step Solution:

1. **Understanding the Unit Circle**:
Since \( a, b, c \) are on the unit circle, we have:
\[ |a| = |b| = |c| = 1 \]

2. **Using the Given Equation**:
We know from the problem statement that:
\[ abc = a + b + c \]

3. **Taking the Modulus**:
Taking the modulus on both sides of the equation \( abc = a + b + c \):
\[ |abc| = |a + b + c| \]
Since \( |abc| = |a| \cdot |b| \cdot |c| = 1 \cdot 1 \cdot 1 = 1 \), we have:
\[ |a + b + c| = 1 \]

4. **Finding \( |ab + bc + ca| \)**:
We need to find \( |ab + ac + bc| \). We can use the identity that relates the modulus of sums and products of complex numbers.

We know that:
\[ |ab + ac + bc| = |a(b+c) + bc| \]
Since \( |a| = 1 \), we can factor out \( a \):
\[ |ab + ac + bc| = |a| \cdot |b+c + bc/a| = |b+c + bc/a| \]

5. **Using the Property of Conjugates**:
Since \( a, b, c \) are on the unit circle, we also have:
\[ \overline{a} = \frac{1}{a}, \quad \overline{b} = \frac{1}{b}, \quad \overline{c} = \frac{1}{c} \]
Thus, we can express \( b+c \) in terms of their conjugates:
\[ |b+c| \leq |b| + |c| = 1 + 1 = 2 \]
However, we need to find \( |ab + ac + bc| \).

6. **Final Calculation**:
By substituting \( b+c = 1 - a \) (from the equation \( a + b + c = 1 \)), we can find:
\[ |ab + ac + bc| = |a(1-a) + bc| \]
Since \( |bc| = 1 \), we can conclude that:
\[ |ab + ac + bc| = |1| = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]