For any three vectors `veca, vecb, vecc` the value of `[(veca+vecb,vecb+vecc,vecc+veca)]` is

Options

A. `0`
B. `2[(veca, vecb, vecc)]`
C. `[(veca, vecb, vecc)]`
D. `-[(veca, vecb, vecc)]`

Explanation

To solve the problem, we need to find the value of the scalar triple product \([( \vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a} )]\).

### Step-by-step Solution:

1. **Define the Vectors**:
Let:
\[ \vec{x} = \vec{a} + \vec{b} \]
\[ \vec{y} = \vec{b} + \vec{c} \]
\[ \vec{z} = \vec{c} + \vec{a} \]

2. **Set Up the Determinant**:
The scalar triple product can be represented as a determinant:
\[ [\vec{x}, \vec{y}, \vec{z}] = \begin{vmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{vmatrix} \]
Here, the columns represent the coefficients of \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) in each vector.

3. **Calculate the Determinant**:
We can calculate the determinant using the formula for a \(3 \times 3\) determinant:
\[ \text{Det} = a(ei - fh) - b(di - fg) + c(dh - eg) \]
For our determinant:
\[ = 1 \cdot (1 \cdot 1 - 1 \cdot 0) - 1 \cdot (0 \cdot 1 - 1 \cdot 1) + 0 \cdot (0 \cdot 0 - 1 \cdot 1) \]
Simplifying this gives:
\[ = 1 \cdot (1 - 0) - 1 \cdot (0 - 1) + 0 \]
\[ = 1 + 1 = 2 \]

4. **Final Result**:
Therefore, the value of the scalar triple product is:
\[ [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 2[\vec{a}, \vec{b}, \vec{c}] \]

### Conclusion:
The final answer is:
\[ 2[\vec{a}, \vec{b}, \vec{c}] \]