If `veca, vecb, vecc` are non-coplanar non-zero vectors, then
`(vecaxxvecb)xx(vecaxxvecc)+(vecbxxvecc)xx(vecbxxveca)+(veccxxveca)xx(veccxxvecb)` is equal to

Options

A. `[(veca, vecb, vecc)]^(2)(veca+vecb+vecc)`
B. `[(veca, vecb, vecc)](veca+vecb+vecc)`
C. `vec0`
D. none of these

Explanation

To solve the problem, we need to evaluate the expression:

\[ (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) + (\vec{b} \times \vec{c}) \times (\vec{b} \times \vec{a}) + (\vec{c} \times \vec{a}) \times (\vec{c} \times \vec{b}) \]

where \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar non-zero vectors.

### Step 1: Evaluate the first term

Using the vector triple product identity:

\[ \vec{p} \times (\vec{q} \times \vec{r}) = (\vec{p} \cdot \vec{r}) \vec{q} - (\vec{p} \cdot \vec{q}) \vec{r} \]

Let \(\vec{p} = \vec{a}\), \(\vec{q} = \vec{b}\), and \(\vec{r} = \vec{c}\). Then we have:

\[ (\vec{a} \times \vec{b}) \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c} \]

### Step 2: Evaluate the second term

Using the same identity for the second term:

\[ (\vec{b} \times \vec{c}) \times (\vec{b} \times \vec{a}) = (\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a} \]

### Step 3: Evaluate the third term

Using the identity for the third term:

\[ (\vec{c} \times \vec{a}) \times (\vec{c} \times \vec{b}) = (\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b} \]

### Step 4: Combine all terms

Now we combine all three results:

\[ [(\vec{a} \cdot \vec{c}) \vec{b} - (\vec{a} \cdot \vec{b}) \vec{c}] + [(\vec{b} \cdot \vec{a}) \vec{c} - (\vec{b} \cdot \vec{c}) \vec{a}] + [(\vec{c} \cdot \vec{b}) \vec{a} - (\vec{c} \cdot \vec{a}) \vec{b}] \]

Grouping like terms, we have:

\[ (\vec{a} \cdot \vec{c}) \vec{b} + (\vec{b} \cdot \vec{a}) \vec{c} + (\vec{c} \cdot \vec{b}) \vec{a} - [(\vec{a} \cdot \vec{b}) \vec{c} + (\vec{b} \cdot \vec{c}) \vec{a} + (\vec{c} \cdot \vec{a}) \vec{b}] \]

### Step 5: Recognizing the result

Notice that the terms can be rearranged and factored as:

\[ \vec{a} \cdot \vec{b} \cdot \vec{c} \text{ (as a scalar triple product)} \]

Thus, the entire expression simplifies to:

\[ \vec{a} \cdot \vec{b} \cdot \vec{c} \]

### Final Result

The final result is:

\[ \vec{a} \cdot \vec{b} \cdot \vec{c} \]