Step 1: Understanding the Concept:
This problem requires applying the vector triple product expansion. The vector triple product formula is $\vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C}$.
Step 2: Key Formula or Approach:
Using the triple product formula:
\[ \hat{i} \times (\vec{a} \times \hat{i}) = (\hat{i} \cdot \hat{i})\vec{a} - (\hat{i} \cdot \vec{a})\hat{i} \]
Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{i} \cdot \vec{a} = a_x$ (the $x$-component of $\vec{a}$):
\[ \hat{i} \times (\vec{a} \times \hat{i}) = \vec{a} - a_x\hat{i} \]
Step 3: Detailed Explanation:
Let the vector $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$.
From our formula:
\[ \hat{i} \times (\vec{a} \times \hat{i}) = \vec{a} - a_x\hat{i} = (a_x\hat{i} + a_y\hat{j} + a_z\hat{k}) - a_x\hat{i} = a_y\hat{j} + a_z\hat{k} \]
The squared magnitude is:
\[ |\hat{i} \times (\vec{a} \times \hat{i})|^2 = |a_y\hat{j} + a_z\hat{k}|^2 = a_y^2 + a_z^2 \]
By symmetry, we can find the other terms:
\[ |\hat{j} \times (\vec{a} \times \hat{j})|^2 = a_x^2 + a_z^2 \]
\[ |\hat{k} \times (\vec{a} \times \hat{k})|^2 = a_x^2 + a_y^2 \]
Now, sum all three parts together:
\[ \text{Sum} = (a_y^2 + a_z^2) + (a_x^2 + a_z^2) + (a_x^2 + a_y^2) \]
\[ \text{Sum} = 2(a_x^2 + a_y^2 + a_z^2) \]
Recognize that $|\vec{a}|^2 = a_x^2 + a_y^2 + a_z^2$, so the expression simplifies to $2|\vec{a}|^2$.
Given $\vec{a} = 2\hat{i} + \hat{j} + 2\hat{k}$, its squared magnitude is:
\[ |\vec{a}|^2 = (2)^2 + (1)^2 + (2)^2 = 4 + 1 + 4 = 9 \]
Substitute this into our simplified sum:
\[ \text{Sum} = 2(9) = 18 \]
Step 4: Final Answer:
The value of the expression is 18. The correct option is (B).