Let vec{a}, vec{b}, vec{c} be three non-zero | vedclass

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(D) Given that $\vec{a} + 3\vec{b}$ is collinear with $\vec{c}$,there exists a scalar $k$ such that $\vec{a} + 3\vec{b} = k\vec{c}$.Also,$\vec{b} + 2\

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$\vec{0}$

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(D) Given that $\vec{a} + 3\vec{b}$ is collinear with $\vec{c}$,there exists a scalar $k$ such that $\vec{a} + 3\vec{b} = k\vec{c}$.Also,$\vec{b} + 2\vec{c}$ is collinear with $\vec{a}$,so there exists a scalar $m$ such that $\vec{b} + 2\vec{c} = m\vec{a}$.From the first equation,$3\vec{b} = k\vec{c} - \vec{a}$,so $\vec{b} = \frac{k}{3}\vec{c} - \frac{1}{3}\vec{a}$.Substitute this into the second equation: $(\frac{k}{3}\vec{c} - \frac{1}{3}\vec{a}) + 2\vec{c} = m\vec{a}$.Rearranging terms: $(\frac{k}{3} + 2)\vec{c} = (m + \frac{1}{3})\vec{a}$.Since $\vec{a}$ and $\vec{c}$ are non-zero and non-collinear,the coefficients must be zero: $\frac{k}{3} + 2 = 0 \implies k = -6$ and $m + \frac{1}{3} = 0 \implies m = -\frac{1}{3}$.Substituting $k = -6$ into the first equation: $\vec{a} + 3\vec{b} = -6\vec{c}$.Therefore,$\vec{a} + 3\vec{b} + 6\vec{c} = \vec{0}$.

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-zero vectors such that no two of them are collinear. If the vector $\vec{a} + 3\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 2\vec{c}$ is collinear with $\vec{a}$,then find the value of $\vec{a} + 3\vec{b} + 6\vec{c}$.

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