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

\(\overline{\mathrm{a}}+2 \overline{\mathrm{b}} \text { is collinear with } \overline{\mathrm{c}} \)
\( \therefore \quad \overline{\mathrm{a}}+2 \overline{\mathrm{b}}=\mathrm{n} \overline{\mathrm{c}} \)
\( \text { Similarly } \overline{\mathrm{b}}+3 \overline{\mathrm{c}}=\mathrm{ma} \)
\( \mathrm{m} \text { and } \mathrm{n} \text { are non-zero scalars. } \)
\( \therefore \text {(i) } \Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}=(\mathrm{n}+6) \overline{\mathrm{c}} \)
\( \text { (ii) } \Rightarrow \overline{\mathrm{a}}+2 \overline{\mathrm{b}}+6 \overline{\mathrm{c}}=(2 \mathrm{~m}+1) \overline{\mathrm{a}} \)
\( \Rightarrow \mathrm{n}+6=0 \text { and } 2 \mathrm{~m}+1=0 \)
\( \Rightarrow \mathrm{n}=-6 \text {and } \mathrm{m}=\frac{-1}{2} \)
\( \therefore \text { (i) } \Rightarrow \mathrm{a}+2 \mathrm{~b}=-6 \overline{\mathrm{c}}\)