Question:

Step 1: Find the relationship between $\vec{a}$ and $\vec{b}$ using the dot product. Taking the dot product of both sides of $\vec{a} = \vec{b} \times \vec{c} + 2\vec{b}$ with $\vec{b}$: $$\vec{a} \cdot \vec{b} = (\vec{b} \times \vec{c}) \cdot \vec{b} + 2\vec{b} \cdot \vec{b}$$ Since the cross product $\vec{b} \times \vec{c}$ is perpendicular to $\vec{b}$, we have $(\vec{b} \times \vec{c}) \cdot \vec{b} = 0$. $$\vec{a} \cdot \vec{b} = 0 + 2|\vec{b}|^2 = 2(4) = 8$$ Step 2: Apply the Cauchy-Schwarz inequality to determine $|\vec{a}|$. By the Cauchy-Schwarz inequality: $$|\vec{a} \cdot \vec{b}| \leq |\vec{a}| \cdot |\vec{b}|$$ Substituting the known values: $$8 \leq |\vec{a}| \cdot 2$$ Given that $|\vec{a}| \leq 4$, we have: $$8 \leq |\vec{a}| \cdot 2 \leq 4 \cdot 2 = 8$$ This equality holds only when $|\vec{a}| = 4$ and $\vec{a}$ is parallel to $\vec{b}$ in the same direction. Therefore, $|\vec{a}| = 4$. Step 3: Determine the relationship between $\vec{b}$ and $\vec{c}$. Since $\vec{a} \parallel \vec{b}$ and $\vec{a} = \vec{b} \times \vec{c} + 2\vec{b}$, the vector $\vec{b} \times \vec{c}$ must also be parallel to $\vec{b}$. However, by definition of the cross product, $\vec{b} \times \vec{c}$ is perpendicular to $\vec{b}$. The only vector that is both parallel and perpendicular to $\vec{b}$ is the zero vector. Therefore: $\vec{b} \times \vec{c} = \vec{0}$, which means $\vec{c} \parallel \vec{b}$. Since $|\vec{c}| = |\vec{b}| = 2$, we have either $\vec{c} = \vec{b}$ or $\vec{c} = -\vec{b}$. Step 4: Evaluate Case 1 where $\vec{c} = \vec{b}$. When $\vec{c} = \vec{b}$: $$\vec{a} = \vec{b} \times \vec{b} + 2\vec{b} = \vec{0} + 2\vec{b} = 2\vec{b}$$ This gives $|\vec{a}| = 2|\vec{b}| = 4$ ✓ Now calculate $|2\vec{a} + \vec{b} + \vec{c}|$: $$|2\vec{a} + \vec{b} + \vec{c}| = |2(2\vec{b}) + \vec{b} + \vec{b}| = |4\vec{b} + \vec{b} + \vec{b}| = |6\vec{b}| = 6 \cdot 2 = 12$$ Step 5: Evaluate Case 2 where $\vec{c} = -\vec{b}$. When $\vec{c} = -\vec{b}$: $$\vec{a} = \vec{b} \times (-\vec{b}) + 2\vec{b} = \vec{0} + 2\vec{b} = 2\vec{b}$$ This gives $|\vec{a}| = 4$ ✓ Now calculate $|2\vec{a} + \vec{b} + \vec{c}|$: $$|2\vec{a} + \vec{b} + \vec{c}| = |2(2\vec{b}) + \vec{b} + (-\vec{b})| = |4\vec{b} + \vec{b} - \vec{b}| = |4\vec{b}| = 4 \cdot 2 = 8$$ Step 6: Find the sum of all possible values. The two possible values of $|2\vec{a} + \vec{b} + \vec{c}|$ are 12 and 8. The sum of possible values is: $$12 + 8 = 20$$ The answer is **Option 3: 20**.