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(D) Given that $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 3\vec{c}$ is collinear with $\vec{a}$.This implies $\vec{a} + 2\vec{b}

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$\vec{0}$

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(D) Given that $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 3\vec{c}$ is collinear with $\vec{a}$.This implies $\vec{a} + 2\vec{b} = k_1 \vec{c}$ and $\vec{b} + 3\vec{c} = k_2 \vec{a}$ for some non-zero scalars $k_1$ and $k_2$.From the first equation,$\vec{a} = k_1 \vec{c} - 2\vec{b}$.Substituting this into the second equation: $\vec{b} + 3\vec{c} = k_2(k_1 \vec{c} - 2\vec{b})$.Rearranging the terms: $\vec{b} + 3\vec{c} = k_1 k_2 \vec{c} - 2k_2 \vec{b}$.$(1 + 2k_2) \vec{b} + (3 - k_1 k_2) \vec{c} = \vec{0}$.Since $\vec{b}$ and $\vec{c}$ are not collinear,their coefficients must be zero:$1 + 2k_2 = 0 \Rightarrow k_2 = -\frac{1}{2}$.$3 - k_1 k_2 = 0 \Rightarrow 3 - k_1(-\frac{1}{2}) = 0 \Rightarrow 3 + \frac{k_1}{2} = 0 \Rightarrow k_1 = -6$.Substituting $k_1 = -6$ into the first equation: $\vec{a} + 2\vec{b} = -6\vec{c}$.Therefore,$\vec{a} + 2\vec{b} + 6\vec{c} = \vec{0}$.

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

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