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

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$10$

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(D) Given that $a \perp (b + c)$,$b \perp (c + a)$,and $c \perp (a + b)$.This implies:$a \cdot (b + c) = 0 \implies a \cdot b + a \cdot c = 0$ .....$(i)$$b \cdot (c + a) = 0 \implies b \cdot c + b \cdot a = 0$ .....$(ii)$$c \cdot (a + b) = 0 \implies c \cdot a + c \cdot b = 0$ .....$(iii)$Adding $(i), (ii),$ and $(iii)$,we get $2(a \cdot b + b \cdot c + c \cdot a) = 0$.Now,consider the sum of the squares of the magnitudes:$|a + b|^2 + |b + c|^2 + |c + a|^2 = 6^2 + 8^2 + 10^2 = 36 + 64 + 100 = 200$.Expanding the left side:$(|a|^2 + |b|^2 + 2a \cdot b) + (|b|^2 + |c|^2 + 2b \cdot c) + (|c|^2 + |a|^2 + 2c \cdot a) = 200$$2(|a|^2 + |b|^2 + |c|^2) + 2(a \cdot b + b \cdot c + c \cdot a) = 200$.Since $2(a \cdot b + b \cdot c + c \cdot a) = 0$,we have $2(|a|^2 + |b|^2 + |c|^2) = 200$,so $|a|^2 + |b|^2 + |c|^2 = 100$.We know that $|a + b + c|^2 = |a|^2 + |b|^2 + |c|^2 + 2(a \cdot b + b \cdot c + c \cdot a)$.Substituting the values: $|a + b + c|^2 = 100 + 0 = 100$.Therefore,$|a + b + c| = \sqrt{100} = 10$.

If $a, b$ and $c$ are perpendicular to $b + c, c + a$ and $a + b$ respectively,and if $|a + b| = 6, |b + c| = 8$ and $|c + a| = 10$,then $|a + b + c| = $

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