We are given that $\overrightarrow a + 3 \overrightarrow b$ is collinear with $\overrightarrow c$, and $\overrightarrow b + 2 \overrightarrow c$ is collinear with $\overrightarrow a$. This means we can write:

1. $\overrightarrow a + 3 \overrightarrow b = \lambda \overrightarrow c \quad ...(i)$
2. $\overrightarrow b + 2 \overrightarrow c = \mu \overrightarrow a \quad ...(ii)$

for some scalars $\lambda$ and $\mu$.

We are trying to find $\overrightarrow a + \overrightarrow b + 6\overrightarrow c$ in terms of $\overrightarrow a$, $\overrightarrow b$, and $\overrightarrow c$. We can also express this as :

$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (\lambda + 6) \overrightarrow c \quad ...(iii)$

by adding $6\overrightarrow c$ to both sides of equation (i).

Now, from equation (ii), multiplying by 3 gives us :

$3\overrightarrow b + 6 \overrightarrow c = 3\mu \overrightarrow a \quad ...(iv)$

Adding $\overrightarrow a$ to both sides of equation (iv) gives :

$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(v)$

Now, we have two expressions for $\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c$, one in terms of $\overrightarrow c$ (from equation iii) and one in terms of $\overrightarrow a$ (from equation v). Setting these equal to each other gives :

$(\lambda + 6) \overrightarrow c = (1 + 3\mu) \overrightarrow a \quad ...(vi)$

Since $\overrightarrow a$ and $\overrightarrow c$ are not collinear, this equation can only hold if the coefficients on both sides are zero, hence :

$\lambda + 6 = 0$ and $1 + 3\mu = 0$

This gives $\lambda = -6$ and $\mu = -\frac{1}{3}$.

Finally, substituting $\lambda = -6$ into equation (iii) gives :

$\overrightarrow a + 3 \overrightarrow b + 6\overrightarrow c = 0$

So, $\overrightarrow a + \overrightarrow b + 6\overrightarrow c = \overrightarrow 0$.

Therefore, the correct answer is Option D : $\overrightarrow 0$.