Step 1: State the given conditions.
$\vec{a}, \vec{b}, \vec{c}$ are unit vectors; $\vec{a}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$; angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$.
Step 2: Use perpendicularity of $\vec{a}$.
Since $\vec{a}$ is perpendicular to the plane of $\vec{b}$ and $\vec{c}$, we have $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \cdot \vec{c} = 0$.
Step 3: Find $|\vec{a}+\vec{b}+\vec{c}|^2$.
$|\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{a}\cdot\vec{c})$.
Step 4: Substitute known values.
$|\vec{a}|^2 = |\vec{b}|^2 = |\vec{c}|^2 = 1$ (unit vectors).
$\vec{a}\cdot\vec{b} = 0$, $\vec{a}\cdot\vec{c} = 0$ (perpendicularity).
$\vec{b}\cdot\vec{c} = |\vec{b}||\vec{c}|\cos\frac{\pi}{3} = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$.
Step 5: Calculate the magnitude squared.
$|\vec{a}+\vec{b}+\vec{c}|^2 = 1+1+1 + 2\left(0 + \frac{1}{2} + 0\right) = 3 + 1 = 4$.
Step 6: Find the magnitude.
$|\vec{a}+\vec{b}+\vec{c}| = \sqrt{4} = 2$. This is option (3).
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