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(D) Given that $\overline{a}+2\overline{b}$ is collinear with $\overline{c}$,we have $\overline{a}+2\overline{b} = n\overline{c}$ for some non-zero sc

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$\overline{0}$

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(D) Given that $\overline{a}+2\overline{b}$ is collinear with $\overline{c}$,we have $\overline{a}+2\overline{b} = n\overline{c}$ for some non-zero scalar $n$. $(i)$Similarly,$\overline{b}+3\overline{c}$ is collinear with $\overline{a}$,so $\overline{b}+3\overline{c} = m\overline{a}$ for some non-zero scalar $m$. (ii)From (ii),we have $\overline{b} = m\overline{a} - 3\overline{c}$.Substitute this into $(i)$: $\overline{a} + 2(m\overline{a} - 3\overline{c}) = n\overline{c}$.This simplifies to $(1+2m)\overline{a} = (n+6)\overline{c}$.Since $\overline{a}$ and $\overline{c}$ are not collinear,the coefficients must be zero: $1+2m = 0 \Rightarrow m = -1/2$ and $n+6 = 0 \Rightarrow n = -6$.Now,consider the expression $\overline{a}+2\overline{b}+6\overline{c}$.From $(i)$,$\overline{a}+2\overline{b} = n\overline{c} = -6\overline{c}$.Therefore,$\overline{a}+2\overline{b}+6\overline{c} = -6\overline{c} + 6\overline{c} = \overline{0}$.

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