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(C) Given that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$.This implies the dot products ar

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$\sqrt{14}$

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(C) Given that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$.This implies the dot products are zero:$\vec a \cdot (\vec b + \vec c) = 0 \Rightarrow \vec a \cdot \vec b + \vec a \cdot \vec c = 0$$\vec b \cdot (\vec c + \vec a) = 0 \Rightarrow \vec b \cdot \vec c + \vec b \cdot \vec a = 0$$\vec c \cdot (\vec a + \vec b) = 0 \Rightarrow \vec c \cdot \vec a + \vec c \cdot \vec b = 0$Adding these equations,we get $2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a) = 0$,which implies $\vec a \cdot \vec b = \vec b \cdot \vec c = \vec c \cdot \vec a = 0$.Now,consider $|\vec a + \vec b + \vec c|^2 = |\vec a|^2 + |\vec b|^2 + |\vec c|^2 + 2(\vec a \cdot \vec b + \vec b \cdot \vec c + \vec c \cdot \vec a)$.Substituting the given magnitudes and the dot products:$|\vec a + \vec b + \vec c|^2 = 1^2 + 2^2 + 3^2 + 2(0) = 1 + 4 + 9 = 14$.Therefore,$|\vec a + \vec b + \vec c| = \sqrt{14}$.

Let $\vec a, \vec b, \vec c$ be three vectors such that $\vec a \perp (\vec b + \vec c)$,$\vec b \perp (\vec c + \vec a)$,and $\vec c \perp (\vec a + \vec b)$. If $|\vec a| = 1, |\vec b| = 2, |\vec c| = 3$,then $|\vec a + \vec b + \vec c| = \dots$

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