`[[overline(a), overline(b)+overline(c), overline(c)+overline(b)+overline(a)]]=`

Options

A. `[[overline(a), overline(b), overline(c)]]`
B. `-[[overline(a), overline(b), overline(c)]]`
C. `0`
D. `2[[overline(a), overline(b), overline(c)]]`

Answer

To solve the given problem involving the scalar triple product of vectors, we will denote the vectors as follows:

Let:
- \(\overline{a} = \mathbf{A}\)
- \(\overline{b} = \mathbf{B}\)
- \(\overline{c} = \mathbf{C}\)

We need to evaluate the scalar triple product:
\[ \mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) \cdot (\mathbf{C} + \mathbf{B} + \mathbf{A}) \]

This can be expressed as the determinant of a matrix formed by these vectors:
\[ \begin{vmatrix} \mathbf{A} & \mathbf{B} + \mathbf{C} & \mathbf{C} + \mathbf{B} + \mathbf{A} \end{vmatrix} \]

### Step 1: Write the determinant
We can write the determinant as:
\[ \begin{vmatrix} \mathbf{A} & \mathbf{B} + \mathbf{C} & \mathbf{C} + \mathbf{B} + \mathbf{A} \end{vmatrix} \]

### Step 2: Apply column operations
We can simplify the determinant using column operations. Specifically, we can subtract the first column from the third column:
\[ \mathbf{C} + \mathbf{B} + \mathbf{A} - \mathbf{A} = \mathbf{B} + \mathbf{C} \]
Thus, the determinant becomes:
\[ \begin{vmatrix} \mathbf{A} & \mathbf{B} + \mathbf{C} & \mathbf{B} + \mathbf{C} \end{vmatrix} \]

### Step 3: Recognize the result
Now, we can see that the third column is identical to the second column. When two columns of a determinant are the same, the value of the determinant is zero:
\[ \begin{vmatrix} \mathbf{A} & \mathbf{B} + \mathbf{C} & \mathbf{B} + \mathbf{C} \end{vmatrix} = 0 \]

### Conclusion
Thus, the scalar triple product evaluates to zero:
\[ \mathbf{A} \cdot (\mathbf{B} + \mathbf{C}) \cdot (\mathbf{C} + \mathbf{B} + \mathbf{A}) = 0 \]

### Final Answer
The correct option is **zero**.

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