If `veca, vecb` are non zero vectors, then `((vecaxxvecb)xxveca).((vecbxxveca)xxvecb)` equals

Options

A. `-(veca.vecb)|(vecaxxvecb)|^2`
B. `|vecaxxvecb|^(2)veca^(2)`
C. `|vecaxxvecb|^(2)vecb^(2)`
D. `(veca.vecb)|vecaxxvecb|^(2)`

Explanation

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

### Step-by-step Solution:

1. **Define the Expression**:
Let \( \vec{A} = (\vec{a} \times \vec{b}) \times \vec{a} \) and \( \vec{B} = (\vec{b} \times \vec{a}) \times \vec{b} \). We need to find \( \vec{A} \cdot \vec{B} \).

2. **Use the Vector Triple Product Identity**:
The vector triple product identity states that \( \vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z}) \vec{y} - (\vec{x} \cdot \vec{y}) \vec{z} \).

3. **Calculate \( \vec{A} \)**:
\[ \vec{A} = (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a}) \vec{b} - (\vec{b} \cdot \vec{a}) \vec{a} \]
Simplifying, we have:
\[ \vec{A} = |\vec{a}|^2 \vec{b} - (\vec{b} \cdot \vec{a}) \vec{a} \]

4. **Calculate \( \vec{B} \)**:
\[ \vec{B} = (\vec{b} \times \vec{a}) \times \vec{b} = (\vec{b} \cdot \vec{b}) \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \]
Simplifying, we have:
\[ \vec{B} = |\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \]

5. **Calculate \( \vec{A} \cdot \vec{B} \)**:
Now we compute the dot product:
\[ \vec{A} \cdot \vec{B} = \left( |\vec{a}|^2 \vec{b} - (\vec{b} \cdot \vec{a}) \vec{a} \right) \cdot \left( |\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b}) \vec{b} \right) \]

6. **Expand the Dot Product**:
Expanding this gives:
\[ = |\vec{a}|^2 |\vec{b}|^2 (\vec{b} \cdot \vec{a}) - |\vec{a}|^2 (\vec{b} \cdot \vec{b}) \vec{a} \cdot \vec{a} - (\vec{b} \cdot \vec{a})^2 \vec{a} \cdot \vec{b} + (\vec{b} \cdot \vec{a})^2 \vec{b} \cdot \vec{b} \]

7. **Combine Like Terms**:
Notice that terms can be rearranged and simplified:
\[ = |\vec{a}|^2 |\vec{b}|^2 (\vec{b} \cdot \vec{a}) - |\vec{a}|^2 |\vec{b}|^2 - (\vec{b} \cdot \vec{a})^2 + (\vec{b} \cdot \vec{a})^2 \]
The last two terms cancel out.

8. **Final Result**:
Thus, we have:
\[ \vec{A} \cdot \vec{B} = |\vec{a}|^2 |\vec{b}|^2 (\vec{b} \cdot \vec{a}) - |\vec{a}|^2 |\vec{b}|^2 = 0 \]

### Conclusion:
The final result is:
\[ \boxed{0} \]