Given:

$$ \vec{a} = 4\hat{i} + 3\hat{j} + \hat{k} $$

$$ \vec{b} = \hat{i} - 2\hat{j} + 2\hat{k} $$

We are asked to find:

$$ \vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b}))) $$

Let's proceed step by step.

Step 1: Compute $\vec{a} \times \vec{b}$

Use the formula for cross product:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ 1 & -2 & 2 \\ \end{vmatrix} $$

Expanding determinant:

$ = \hat{i} \begin{vmatrix} 3 & 1 \\ -2 & 2 \\ \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 1 \\ 1 & 2 \\ \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 3 \\ 1 & -2 \\ \end{vmatrix} $

$ = \hat{i}[3 \times 2 - 1 \times (-2)] - \hat{j}[4 \times 2 - 1 \times 1] + \hat{k}[4 \times (-2) - 3 \times 1] $

$ = \hat{i}[6 + 2] - \hat{j}[8 - 1] + \hat{k}[-8 - 3] $

$ = 8\hat{i} - 7\hat{j} - 11\hat{k} $

Step 2: Compute $\vec{a} \times (\vec{a} \times \vec{b})$

Let $\vec{c} = \vec{a} \times \vec{b} = 8\hat{i} - 7\hat{j} - 11\hat{k}$

Now,

$ \vec{a} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ 8 & -7 & -11 \\ \end{vmatrix} $

Expanding:

$ = \hat{i} \begin{vmatrix} 3 & 1 \\ -7 & -11 \\ \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 1 \\ 8 & -11 \\ \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 3 \\ 8 & -7 \\ \end{vmatrix} $

$ = \hat{i}[3 \times (-11) - 1 \times (-7)] - \hat{j}[4 \times (-11) - 1 \times 8] + \hat{k}[4 \times (-7) - 3 \times 8] $

$ = \hat{i}[-33 + 7] - \hat{j}[-44 - 8] + \hat{k}[-28 - 24] $

$ = -26\hat{i} + 52\hat{j} - 52\hat{k} $

Step 3: Compute $\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b}))$

Let $\vec{d} = -26\hat{i} + 52\hat{j} - 52\hat{k}$

$ \vec{a} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -26 & 52 & -52 \\ \end{vmatrix} $

Expanding:

$ = \hat{i} \begin{vmatrix} 3 & 1 \\ 52 & -52 \\ \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 1 \\ -26 & -52 \\ \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 3 \\ -26 & 52 \\ \end{vmatrix} $

$ = \hat{i}[3 \times (-52) - 1 \times 52] - \hat{j}[4 \times (-52) - 1 \times (-26)] + \hat{k}[4 \times 52 - 3 \times (-26)] $

$ = \hat{i}[-156 - 52] - \hat{j}[-208 + 26] + \hat{k}[208 + 78] $

$ = -208\hat{i} + 182\hat{j} + 286\hat{k} $

Step 4: Compute $\vec{a} \times (\vec{a} \times (\vec{a} \times (\vec{a} \times \vec{b})))$

Let $\vec{e} = -208\hat{i} + 182\hat{j} + 286\hat{k}$

$ \vec{a} \times \vec{e} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 3 & 1 \\ -208 & 182 & 286 \\ \end{vmatrix} $

Expanding:

$ = \hat{i} \begin{vmatrix} 3 & 1 \\ 182 & 286 \\ \end{vmatrix} - \hat{j} \begin{vmatrix} 4 & 1 \\ -208 & 286 \\ \end{vmatrix} + \hat{k} \begin{vmatrix} 4 & 3 \\ -208 & 182 \\ \end{vmatrix} $

$ = \hat{i}[3 \times 286 - 1 \times 182] - \hat{j}[4 \times 286 - 1 \times (-208)] + \hat{k}[4 \times 182 - 3 \times (-208)] $

$ = \hat{i}[858 - 182] - \hat{j}[1144 + 208] + \hat{k}[728 + 624] $

$ = \hat{i}[676] - \hat{j}[1352] + \hat{k}[1352] $

$ = 676\hat{i} - 1352\hat{j} + 1352\hat{k} $

Now, observe that

$ \vec{b} = \hat{i} - 2\hat{j} + 2\hat{k} $

$ 676\vec{b} = 676\hat{i} - 1352\hat{j} + 1352\hat{k} $

So, the answer is

$ \boxed{676\;\vec{b}} $

This matches Option B.

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