Three non-zero non-collinear vectors vec{a}, | vedclass

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(A) Given that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$.Therefore,$\vec{a}+3\vec{b} = \lambda\vec{c}$ for some scalar $\lambda$.This implies $\v

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(A) Given that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$.Therefore,$\vec{a}+3\vec{b} = \lambda\vec{c}$ for some scalar $\lambda$.This implies $\vec{a}+3\vec{b}-\lambda\vec{c} = 0$  $(i)$.Also,$3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$.Therefore,$3\vec{b}+2\vec{c} = \mu\vec{a}$ for some scalar $\mu$.This implies $3\vec{b}+2\vec{c}-\mu\vec{a} = 0$  $(ii)$.From $(i)$,we have $3\vec{b} = \lambda\vec{c} - \vec{a}$.Substituting this into $(ii)$:$(\lambda\vec{c} - \vec{a}) + 2\vec{c} - \mu\vec{a} = 0$$(\lambda+2)\vec{c} - (1+\mu)\vec{a} = 0$.Since $\vec{a}$ and $\vec{c}$ are non-collinear,their coefficients must be zero.Thus,$\lambda+2 = 0 \implies \lambda = -2$ and $1+\mu = 0 \implies \mu = -1$.Substituting $\lambda = -2$ into $(i)$:$\vec{a}+3\vec{b} - (-2)\vec{c} = 0$$\vec{a}+3\vec{b}+2\vec{c} = 0$.

Three non-zero non-collinear vectors $\vec{a}, \vec{b}$ and $\vec{c}$ are such that $\vec{a}+3\vec{b}$ is collinear with $\vec{c}$,and $3\vec{b}+2\vec{c}$ is collinear with $\vec{a}$. Then $\vec{a}+3\vec{b}+2\vec{c}$ equals to

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