`vecA, vecB and vecC` are vectors each having a unit magneitude. If `vecA + vecB+vecC=0`, then `vecA. vecB+vecB.vecC+ vecC.vecA` will be:

Options

A. 1
B. `-(3)/(2)`
C. `-(1)/(2)`
D. 0

Answer

To solve the problem, we start with the given information that vectors \(\vec{A}\), \(\vec{B}\), and \(\vec{C}\) each have unit magnitude and that:

\[ \vec{A} + \vec{B} + \vec{C} = 0 \]

### Step 1: Rearranging the Equation
From the equation \(\vec{A} + \vec{B} + \vec{C} = 0\), we can rearrange it to express one vector in terms of the others:

\[ \vec{C} = -(\vec{A} + \vec{B}) \]

### Step 2: Squaring Both Sides
Next, we will square the magnitude of both sides of the equation:

\[ |\vec{A} + \vec{B} + \vec{C}|^2 = 0 \]

Expanding the left side using the dot product:

\[ |\vec{A}|^2 + |\vec{B}|^2 + |\vec{C}|^2 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0 \]

### Step 3: Substituting Magnitudes
Since each vector has unit magnitude, we have:

\[ |\vec{A}|^2 = 1, \quad |\vec{B}|^2 = 1, \quad |\vec{C}|^2 = 1 \]

Substituting these values into the equation:

\[ 1 + 1 + 1 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0 \]

This simplifies to:

\[ 3 + 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = 0 \]

### Step 4: Isolating the Dot Products
Now, we isolate the dot products:

\[ 2(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}) = -3 \]

Dividing both sides by 2 gives:

\[ \vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A} = -\frac{3}{2} \]

### Final Answer
Thus, the value of \(\vec{A} \cdot \vec{B} + \vec{B} \cdot \vec{C} + \vec{C} \cdot \vec{A}\) is:

\[ \boxed{-\frac{3}{2}} \]