Step 1: Understanding the Question:
This is a complex vector expression involving dot products, cross products, and vector triple products. We will simplify it using vector identities.
Step 3: Detailed Explanation:
1. First, check properties of $\vec{a}$ and $\vec{b}$:
$|\vec{a}|^2 = \frac{1}{10}(9+1) = 1 \implies |\vec{a}| = 1$.
$|\vec{b}|^2 = \frac{1}{49}(4+9+36) = 1 \implies |\vec{b}| = 1$.
$\vec{a} \cdot \vec{b} = \frac{1}{7\sqrt{10}} (3(2) + 0(3) + 1(-6)) = 0$.
So $\vec{a}$ and $\vec{b}$ are orthogonal unit vectors.
2. Simplify the vector triple product in the curly brackets:
$\vec{V} = (\vec{a} \times \vec{b}) \times (2\vec{a} + \vec{b})$
Using $\vec{X} \times \vec{Y} \times \vec{Z} = (\vec{X} \cdot \vec{Z})\vec{Y} - (\vec{X} \cdot \vec{Y})\vec{Z}$? No, use $(\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$:
$\vec{V} = [\vec{a} \cdot (2\vec{a} + \vec{b})]\vec{b} - [\vec{b} \cdot (2\vec{a} + \vec{b})]\vec{a}$
$\vec{V} = [2|\vec{a}|^2 + \vec{a} \cdot \vec{b}]\vec{b} - [2\vec{b} \cdot \vec{a} + |\vec{b}|^2]\vec{a}$
$\vec{V} = [2(1) + 0]\vec{b} - [0 + 1]\vec{a} = 2\vec{b} - \vec{a}$.
3. Calculate the total expression:
$Result = (\vec{a} - 2\vec{b}) \cdot (2\vec{b} - \vec{a})$
$Result = -(\vec{a} - 2\vec{b}) \cdot (\vec{a} - 2\vec{b}) = -|\vec{a} - 2\vec{b}|^2$.
4. Since $\vec{a} \perp \vec{b}$:
$|\vec{a} - 2\vec{b}|^2 = |\vec{a}|^2 + |-2\vec{b}|^2 = 1 + 4(1) = 5$.
So the final value is $-5$.
Step 4: Final Answer:
The value of the expression is $-5$.