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(D) Given that $a \cdot (b + c) = 0$, $b \cdot (c + a) = 0$, and $c \cdot (a + b) = 0$.This implies $a \cdot b + a \cdot c = 0$, $b \cdot c + b \cdot

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$\sqrt{14}$

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(D) Given that $a \cdot (b + c) = 0$, $b \cdot (c + a) = 0$, and $c \cdot (a + b) = 0$.This implies $a \cdot b + a \cdot c = 0$, $b \cdot c + b \cdot a = 0$, and $c \cdot a + c \cdot b = 0$.Adding these three equations: $2(a \cdot b + b \cdot c + c \cdot a) = 0$, so $a \cdot b + b \cdot c + c \cdot a = 0$.Since $a \cdot b + a \cdot c = 0$, it follows that $a \cdot b = 0$, $b \cdot c = 0$, and $c \cdot a = 0$.Now, $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.$|a + b + c|^2 = 1^2 + 2^2 + 3^2 + 2(0) = 1 + 4 + 9 = 14$.Therefore, $|a + b + c| = \sqrt{14}$.

If $a, b, c$ are three vectors such that $a \perp (b + c)$, $b \perp (c + a)$ and $c \perp (a + b)$, and $|a| = 1, |b| = 2, |c| = 3$, then $|a + b + c|$ is equal to:

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