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(D) Given that $a$,$b$,and $c$ are non-zero vectors such that no two of these are collinear. Since $a + 2b$ is collinear with $c$,we have $a + 2b = m

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(D) Given that $a$,$b$,and $c$ are non-zero vectors such that no two of these are collinear. Since $a + 2b$ is collinear with $c$,we have $a + 2b = m c$ for some non-zero scalar $m$ $(i)$. Since $b + 3c$ is collinear with $a$,we have $b + 3c = n a$ for some non-zero scalar $n$ $(ii)$. From $(ii)$,we have $b = n a - 3c$. Substitute this into $(i)$: $a + 2(n a - 3c) = m c$ $a + 2n a - 6c = m c$ $(1 + 2n) a = (m + 6) c$ Since $a$ and $c$ are non-collinear,the coefficients must be zero: $1 + 2n = 0 \Rightarrow n = -\frac{1}{2}$ $m + 6 = 0 \Rightarrow m = -6$ Substituting $m = -6$ into equation $(i)$: $a + 2b = -6c$ $a + 2b + 6c = 0$.

Let $a$,$b$,and $c$ be $3$ non-zero vectors such that no $2$ of these are collinear. If vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then $a + 2b + 6c$ equals:

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