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If $\bar{a}, \bar{b}, \bar{c}$ are three non-zero vectors, no two of them are collinear, $\bar{a}+2 \bar{b}$ is collinear with $\overline{\mathrm{c}}, \overline{\mathrm{b}}+3 \overline{\mathrm{c}}$ is collinear with $\overline{\mathrm{a}}$, then $\overline{\mathrm{a}}+2 \overline{\mathrm{b}}$ is
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$-6 \bar{c}$
$\begin{array}{ll} & \overline{\mathrm{a}}+2 \overline{\mathrm{b}} \text { is collinear with } \overline{\mathrm{c}} \\ \therefore \quad & \overline{\mathrm{a}}+2 \overline{\mathrm{b}}=\mathrm{n} \overline{\mathrm{c}} \\ & \text { Similarly } \overline{\mathrm{b}}+3 \overline{\mathrm{c}}=\mathrm{ma} \\ & \mathrm{m} \text { and } \mathrm{n} \text { are non-zero scalars. } \\ \therefore \quad & \text { (i) } \Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}=(\mathrm{n}+6) \overline{\mathrm{c}} \\ & \text { (ii) } \Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}=(2 \mathrm{~m}+1) \overline{\mathrm{a}} \\ \Rightarrow \mathrm{n}+6=0 \text { and } 2 \mathrm{~m}+1=0 \\ \Rightarrow \mathrm{n}=-6 \text { and } \mathrm{m}=\frac{-1}{2} \\ \therefore \quad & \text { (i) } \Rightarrow \mathrm{a}+2 \mathrm{~b}=-6 \overline{\mathrm{c}}\end{array}$
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