For an vector `veca` the value of
`hatixx(vecaxxveci)+hatjxx(vecaxxhatj)+hatkxx(vecaxxveck)`, is

Options

A. `veca`
B. `2veca`
C. `3veca`
D. `vec0`

Explanation

To solve the given expression \( \hat{i} \times (\vec{a} \times \hat{i}) + \hat{j} \times (\vec{a} \times \hat{j}) + \hat{k} \times (\vec{a} \times \hat{k}) \), we will use the vector triple product identity, which states that \( \vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w}) \vec{v} - (\vec{u} \cdot \vec{v}) \vec{w} \).

### Step-by-step Solution:

1. **Identify the vectors**:
Let \( \vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k} \).

2. **Calculate \( \hat{i} \times (\vec{a} \times \hat{i}) \)**:
Using the vector triple product identity:
\[ \hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \cdot \hat{i}) \vec{a} - (\hat{i} \cdot \vec{a}) \hat{i} \]
Here, \( \hat{i} \cdot \hat{i} = 1 \) and \( \hat{i} \cdot \vec{a} = a_1 \).
Thus,
\[ \hat{i} \times (\vec{a} \times \hat{i}) = \vec{a} - a_1 \hat{i} = (a_2 \hat{j} + a_3 \hat{k}). \]

3. **Calculate \( \hat{j} \times (\vec{a} \times \hat{j}) \)**:
Similarly,
\[ \hat{j} \times (\vec{a} \times \hat{j}) = (\hat{j} \cdot \hat{j}) \vec{a} - (\hat{j} \cdot \vec{a}) \hat{j} \]
Here, \( \hat{j} \cdot \hat{j} = 1 \) and \( \hat{j} \cdot \vec{a} = a_2 \).
Thus,
\[ \hat{j} \times (\vec{a} \times \hat{j}) = \vec{a} - a_2 \hat{j} = (a_1 \hat{i} + a_3 \hat{k}). \]

4. **Calculate \( \hat{k} \times (\vec{a} \times \hat{k}) \)**:
Following the same process,
\[ \hat{k} \times (\vec{a} \times \hat{k}) = (\hat{k} \cdot \hat{k}) \vec{a} - (\hat{k} \cdot \vec{a}) \hat{k} \]
Here, \( \hat{k} \cdot \hat{k} = 1 \) and \( \hat{k} \cdot \vec{a} = a_3 \).
Thus,
\[ \hat{k} \times (\vec{a} \times \hat{k}) = \vec{a} - a_3 \hat{k} = (a_1 \hat{i} + a_2 \hat{j}). \]

5. **Combine all results**:
Now, we can combine the results from steps 2, 3, and 4:
\[ \hat{i} \times (\vec{a} \times \hat{i}) + \hat{j} \times (\vec{a} \times \hat{j}) + \hat{k} \times (\vec{a} \times \hat{k}) = (a_2 \hat{j} + a_3 \hat{k}) + (a_1 \hat{i} + a_3 \hat{k}) + (a_1 \hat{i} + a_2 \hat{j}). \]
Simplifying this gives:
\[ = 2a_1 \hat{i} + 2a_2 \hat{j} + 2a_3 \hat{k} = 2(a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}) = 2\vec{a}. \]

### Final Answer:
Thus, the value of the expression is \( 2\vec{a} \).