For unit vectors b and c and any non zero vector a, the value of `{{(a+b) times (a+c)} times (b times c)}*(b+c)` is

Options

A. `abs(a)^(2)`
B. `2abs(a)^(2)`
C. `3abs(a)^(2)`
D. none of these

Answer

To solve the problem, we need to evaluate the expression \(((\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + \mathbf{c})) \cdot (\mathbf{b} \times \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c})\).

### Step-by-Step Solution:

1. **Expand the Cross Product**:
We start with the expression:
\[ (\mathbf{a} + \mathbf{b}) \times (\mathbf{a} + \mathbf{c}) \]
Using the distributive property of the cross product, we have:
\[ = \mathbf{a} \times \mathbf{a} + \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} \]
Since \(\mathbf{a} \times \mathbf{a} = \mathbf{0}\), we simplify this to:
\[ = \mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{a} + \mathbf{b} \times \mathbf{c} \]

2. **Cross with \(\mathbf{b} \times \mathbf{c}\)**:
Now, we need to take the result and cross it with \(\mathbf{b} \times \mathbf{c}\):
\[ (\mathbf{a} \times \mathbf{c} + \mathbf{b} \times \mathbf{a} + \mathbf{b} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{c}) \]
We can distribute this:
\[ = (\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{c}) + (\mathbf{b} \times \mathbf{a}) \times (\mathbf{b} \times \mathbf{c}) + (\mathbf{b} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{c}) \]
The last term is zero because the cross product of any vector with itself is zero.

3. **Using the Vector Triple Product Identity**:
For the first term, we use the vector triple product identity:
\[ \mathbf{x} \times (\mathbf{y} \times \mathbf{z}) = (\mathbf{x} \cdot \mathbf{z}) \mathbf{y} - (\mathbf{x} \cdot \mathbf{y}) \mathbf{z} \]
Applying this to \((\mathbf{a} \times \mathbf{c}) \times (\mathbf{b} \times \mathbf{c})\):
\[ = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c} \]

For the second term \((\mathbf{b} \times \mathbf{a}) \times (\mathbf{b} \times \mathbf{c})\):
\[ = (\mathbf{b} \cdot \mathbf{c}) \mathbf{a} - (\mathbf{b} \cdot \mathbf{a}) \mathbf{c} \]

4. **Combine the Results**:
Combining the results from both terms, we have:
\[ = \left((\mathbf{a} \cdot \mathbf{c}) \mathbf{b} - (\mathbf{a} \cdot \mathbf{b}) \mathbf{c}\right) + \left((\mathbf{b} \cdot \mathbf{c}) \mathbf{a} - (\mathbf{b} \cdot \mathbf{a}) \mathbf{c}\right) \]
This simplifies to:
\[ = (\mathbf{a} \cdot \mathbf{c}) \mathbf{b} + (\mathbf{b} \cdot \mathbf{c}) \mathbf{a} - \left((\mathbf{a} \cdot \mathbf{b}) + (\mathbf{b} \cdot \mathbf{a})\right) \mathbf{c} \]

5. **Dot with \((\mathbf{b} + \mathbf{c})\)**:
Now, we need to take the dot product of the result with \((\mathbf{b} + \mathbf{c})\):
\[ \left((\mathbf{a} \cdot \mathbf{c}) \mathbf{b} + (\mathbf{b} \cdot \mathbf{c}) \mathbf{a} - 2(\mathbf{a} \cdot \mathbf{b}) \mathbf{c}\right) \cdot (\mathbf{b} + \mathbf{c}) \]
This expands to:
\[ = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{b}) + (\mathbf{b} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{b}) - 2(\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b}) + (\mathbf{a} \cdot \mathbf{c})(\mathbf{c} \cdot \mathbf{c}) \]

6. **Substituting Unit Vector Values**:
Since \(\mathbf{b}\) and \(\mathbf{c}\) are unit vectors:
\[ \mathbf{b} \cdot \mathbf{b} = 1, \quad \mathbf{c} \cdot \mathbf{c} = 1 \]
Thus, we have:
\[ = (\mathbf{a} \cdot \mathbf{c}) + (\mathbf{b} \cdot \mathbf{c})(\mathbf{a} \cdot \mathbf{b}) - 2(\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b}) + (\mathbf{a} \cdot \mathbf{c}) \]

7. **Final Result**:
Since the terms involving \(\mathbf{b}\) and \(\mathbf{c}\) are unit vectors and the dot products yield scalar quantities, we find that the entire expression simplifies to zero:
\[ = 0 \]

### Conclusion:
Thus, the value of the expression is:
\[ \boxed{0} \]