Let $$\vec{a}, \vec{b}$$ and $$\vec{c}$$ be three non-zero vectors such that no two of these are collinear. If the vector $$\vec{a} + 2\vec{b}$$ is collinear with $$\vec{c}$$ and $$\vec{b} + 3\vec{c}$$ is collinear with $$\vec{a}$$ ($$\lambda$$ being some non-zero scalar) then $$\vec{a} + 2\vec{b} + 6\vec{c}$$ equals

Since

$$\vec a+2\vec b$$ is collinear with $$\vec c$$ there exists a scalar $$m$$ such that $$\vec a+2\vec b=m\vec c$$

Therefore,

$$\vec a+2\vec b-m\vec c=\vec 0 \qquad (1)$$

Also,

$$\vec b+3\vec c$$ is collinear with $$\vec a$$

Hence there exists a scalar $$n$$ such that

$$\vec b+3\vec c=n\vec a$$

Therefore,

$$-n\vec a+\vec b+3\vec c=\vec 0 \qquad (2)$$

Since no two of $$\vec a,\vec b,\vec c$$ are collinear, they cannot satisfy two independent linear relations.

Hence equations (1) and (2) must be proportional.

Comparing coefficients,

$$\frac{-n}{1}=\frac{1}{2}=\frac{3}{-m}$$

From

$$\frac{1}{2}=\frac{-n}{1}$$

we get

$$n=-\frac12$$

and from

$$\frac{1}{2}=\frac{3}{-m}$$

we get

$$m=-6$$

Substituting $$m=-6$$ into

$$\vec a+2\vec b=m\vec c$$ gives $$\vec a+2\vec b=-6\vec c$$

Therefore,

$$\vec a+2\vec b+6\vec c=\vec 0$$

Hence,

$$\boxed{\vec 0}$$