If `veca , vecb ,vec c ` are the 3 vectors such that `|veca| = 3, |vecb| = 4,|vec c| = 5, |veca + vecb + vec c | = 0` then the value of `veca.vecb + vecb.vec c + vec c .vec a ` is :

Options

A. -20
B. -25
C. 25
D. 50

Answer

To solve the problem, we need to find the value of \( \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} \) given the magnitudes of the vectors and the condition that \( |\vec{a} + \vec{b} + \vec{c}| = 0 \).

### Step-by-Step Solution:

1. **Understanding the Condition**:
Since \( |\vec{a} + \vec{b} + \vec{c}| = 0 \), it implies that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \). This means that the vectors \( \vec{a}, \vec{b}, \vec{c} \) form a closed triangle.

2. **Using the Dot Product**:
We can take the dot product of \( \vec{a} + \vec{b} + \vec{c} \) with itself:
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = 0 \]
Expanding this, we have:
\[ (\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0 \]
This expands to:
\[ \vec{a} \cdot \vec{a} + \vec{b} \cdot \vec{b} + \vec{c} \cdot \vec{c} + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

3. **Substituting the Magnitudes**:
We know the magnitudes:
\[ |\vec{a}| = 3 \implies \vec{a} \cdot \vec{a} = 3^2 = 9 \]
\[ |\vec{b}| = 4 \implies \vec{b} \cdot \vec{b} = 4^2 = 16 \]
\[ |\vec{c}| = 5 \implies \vec{c} \cdot \vec{c} = 5^2 = 25 \]
Substituting these values into the equation gives:
\[ 9 + 16 + 25 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]

4. **Simplifying the Equation**:
Combine the constants:
\[ 50 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0 \]
Rearranging gives:
\[ 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = -50 \]

5. **Finding the Required Value**:
Dividing both sides by 2:
\[ \vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a} = -25 \]

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