The three vectors a, b and c with magnitude 3, 4 and 5 respectively and `a+b+c=0,` then the value of `a.b+b.c+c.a` is

Options

A. `-23`
B. `-25`
C. 30
D. 26

Explanation

To solve the problem, we need to find the value of \( a \cdot b + b \cdot c + c \cdot a \) given that the vectors \( a, b, c \) have magnitudes 3, 4, and 5 respectively, and that \( a + b + c = 0 \).

### Step-by-Step Solution:

1. **Understanding the Equation**:
Since \( a + b + c = 0 \), we can rearrange this to \( c = - (a + b) \).

2. **Squaring the Equation**:
We can square both sides of the equation \( a + b + c = 0 \):
\[ |a + b + c|^2 = 0 \]

3. **Expanding the Left Side**:
The square of a vector can be expanded as:
\[ |a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a) \]
Here, \( |a|^2 = 3^2 = 9 \), \( |b|^2 = 4^2 = 16 \), and \( |c|^2 = 5^2 = 25 \).

4. **Substituting the Magnitudes**:
Substitute the magnitudes into the equation:
\[ 0 = 9 + 16 + 25 + 2(a \cdot b + b \cdot c + c \cdot a) \]

5. **Calculating the Sum of Squares**:
Calculate \( 9 + 16 + 25 \):
\[ 9 + 16 = 25 \]
\[ 25 + 25 = 50 \]
Thus, we have:
\[ 0 = 50 + 2(a \cdot b + b \cdot c + c \cdot a) \]

6. **Isolating the Dot Product Terms**:
Rearranging gives:
\[ 2(a \cdot b + b \cdot c + c \cdot a) = -50 \]

7. **Dividing by 2**:
Divide both sides by 2:
\[ a \cdot b + b \cdot c + c \cdot a = -25 \]

### Final Answer:
Thus, the value of \( a \cdot b + b \cdot c + c \cdot a \) is \( \boxed{-25} \).