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(D) Given that $a + 2b$ is collinear with $c$,there exists a scalar $x$ such that $a + 2b = xc$. Given that $b + 3c$ is collinear with $a$,there exist

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(D) Given that $a + 2b$ is collinear with $c$,there exists a scalar $x$ such that $a + 2b = xc$. Given that $b + 3c$ is collinear with $a$,there exists a scalar $y$ such that $b + 3c = ya$. From the first equation,$a + 2b = xc$. From the second equation,$b = ya - 3c$. Substitute $b$ into the first equation: $a + 2(ya - 3c) = xc$. $a + 2ya - 6c = xc$. $(1 + 2y)a = (x + 6)c$. Since $a$ and $c$ are non-zero and non-collinear,the coefficients must be zero: $1 + 2y = 0 \implies y = -\frac{1}{2}$. $x + 6 = 0 \implies x = -6$. Substituting $x = -6$ into $a + 2b = xc$,we get $a + 2b = -6c$,which implies $a + 2b + 6c = 0$.

If $a, b$ and $c$ are three non-zero vectors,no two of which are collinear. If the vector $a + 2b$ is collinear with $c$ and $b + 3c$ is collinear with $a$,then ($\lambda$ being some non-zero scalar) $a + 2b + 6c$ is equal to

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