Let vec{alpha}, vec{beta}, vec{gamma} be thre | vedclass

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(B) Given that $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$, there exists a scalar $k_{1}$ such that $\vec{\alpha}+3 \vec{\beta}=k_{1

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$\overrightarrow{0}$

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(B) Given that $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$, there exists a scalar $k_{1}$ such that $\vec{\alpha}+3 \vec{\beta}=k_{1} \vec{\gamma}$.This implies $\vec{\beta}=\frac{k_{1}}{3} \vec{\gamma}-\frac{1}{3} \vec{\alpha}$.Also, $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, so there exists a scalar $k_{2}$ such that $\vec{\beta}+2 \vec{\gamma}=k_{2} \vec{\alpha}$.This implies $\vec{\beta}=k_{2} \vec{\alpha}-2 \vec{\gamma}$.Equating the two expressions for $\vec{\beta}$, we get $\frac{k_{1}}{3} \vec{\gamma}-\frac{1}{3} \vec{\alpha}=k_{2} \vec{\alpha}-2 \vec{\gamma}$.Rearranging the terms, we have $\vec{\alpha}(k_{2}+\frac{1}{3})=\vec{\gamma}(\frac{k_{1}}{3}+2)$.Since $\vec{\alpha}$ and $\vec{\gamma}$ are non-collinear, the coefficients must be zero: $k_{2}+\frac{1}{3}=0 \Rightarrow k_{2}=-\frac{1}{3}$ and $\frac{k_{1}}{3}+2=0 \Rightarrow k_{1}=-6$.Substituting $k_{1}=-6$ into the first equation, we get $\vec{\alpha}+3 \vec{\beta}=-6 \vec{\gamma}$.Therefore, $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}$.

Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$ and $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, then $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}$ is

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