Step 1: Understanding the Concept:
We have a repeated cross product. We should use the vector triple product formula to simplify it step by step from the inside out.
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:
Apply the formula to the innermost triple product: $\bar{a} \times (\bar{a} \times \vec{v})$, where $\vec{v} = \bar{a} \times \bar{b}$.
Notice that a dot product might simplify to zero, making subsequent steps very easy.
Step 3: Detailed Explanation:
First, let's calculate the dot product $\bar{a} \cdot \bar{b}$ and the magnitude squared $|\bar{a}|^2$.
$\bar{a} = \langle 4, 3, 1 \rangle$ and $\bar{b} = \langle 1, -2, 2 \rangle$.
$\bar{a} \cdot \bar{b} = (4)(1) + (3)(-2) + (1)(2) = 4 - 6 + 2 = 0$.
Since $\bar{a} \cdot \bar{b} = 0$, the vectors $\bar{a}$ and $\bar{b}$ are perpendicular.
$|\bar{a}|^2 = \bar{a} \cdot \bar{a} = 4^2 + 3^2 + 1^2 = 16 + 9 + 1 = 26$.
Let's evaluate the expression from the inside:
Let $\vec{v}_1 = \bar{a} \times \bar{b}$.
Next term is $\vec{v}_2 = \bar{a} \times \vec{v}_1 = \bar{a} \times (\bar{a} \times \bar{b})$.
Using the vector triple product formula:
$\bar{a} \times (\bar{a} \times \bar{b}) = (\bar{a} \cdot \bar{b})\bar{a} - (\bar{a} \cdot \bar{a})\bar{b}$.
Since $\bar{a} \cdot \bar{b} = 0$, this simplifies to:
$\vec{v}_2 = 0\bar{a} - |\bar{a}|^2\bar{b} = -26\bar{b}$.
Now, the next term is $\vec{v}_3 = \bar{a} \times \vec{v}_2 = \bar{a} \times (-26\bar{b}) = -26(\bar{a} \times \bar{b})$.
Finally, evaluate the entire expression:
$\bar{a} \times \vec{v}_3 = \bar{a} \times (-26(\bar{a} \times \bar{b})) = -26 [\bar{a} \times (\bar{a} \times \bar{b})]$.
We already found that $\bar{a} \times (\bar{a} \times \bar{b}) = -26\bar{b}$.
So, substituting this back:
Result $= -26 (-26\bar{b}) = (-26)^2 \bar{b} = 676\bar{b}$.
Step 4: Final Answer:
The result is $676\bar{b}$.