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

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$\sqrt{2}$

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(B) Given that $a, b, c$ are unit vectors, so $|a| = |b| = |c| = 1$.Since $a$ is perpendicular to the plane of $b$ 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, $|a + b + c|^2 = (a + b + c) \cdot (a + b + c) = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.Substituting the values: $|a + b + c|^2 = 1^2 + 1^2 + 1^2 + 2(0 + \frac{1}{2} + 0) = 3 + 1 = 4$.Therefore, $|a + b + c| = \sqrt{4} = 2$.

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

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