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(C) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$. Since $a$ is perpendicular to the plane containing $b$ and $c$,$a \cdot b = 0$ and

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$2$

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(C) Given that $a, b, c$ are unit vectors,so $|a| = |b| = |c| = 1$. Since $a$ is perpendicular to the plane containing $b$ and $c$,$a \cdot b = 0$ and $a \cdot c = 0$. The angle between $b$ and $c$ is $\frac{\pi}{3}$,so $b \cdot c = |b||c| \cos(\frac{\pi}{3}) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$. Now,consider $|a+b+c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$. Substituting the values: $|a+b+c|^2 = 1^2 + 1^2 + 1^2 + 2(0 + \frac{1}{2} + 0)$. $|a+b+c|^2 = 1 + 1 + 1 + 1 = 4$. Therefore,$|a+b+c| = \sqrt{4} = 2$.

Let $a, b$ and $c$ be unit vectors such that $a$ is perpendicular to the plane containing $b$ and $c$ and the angle between $b$ and $c$ is $\frac{\pi}{3}$. Then,$|a+b+c|=$

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