1. Let

$$\vec a \times \vec b = \vec b \times \vec c = \vec c \times \vec a.$$

We need to find $\vec a+\vec b+\vec c$.

2. Since

$$\vec a \times \vec b = \vec b \times \vec c,$$

subtracting gives

$$\vec a \times \vec b - \vec b \times \vec c = 0.$$

Using distributivity of cross product,

$$\vec a \times \vec b + \vec c \times \vec b = 0$$

because $-\vec b\times\vec c=\vec c\times\vec b$. So,

$$(\vec a+\vec c)\times \vec b=0.$$

Hence $\vec a+\vec c$ is parallel to $\vec b$.

3. Also, from

$$\vec b \times \vec c = \vec c \times \vec a,$$

we get

$$\vec b \times \vec c - \vec c \times \vec a=0.$$

Since $-\vec c\times\vec a=\vec a\times\vec c$,

$$\vec b\times\vec c+\vec a\times\vec c=0$$ $$(\vec a+\vec b)\times\vec c=0.$$

Hence $\vec a+\vec b$ is parallel to $\vec c$.

4. Similarly, from

$$\vec c \times \vec a = \vec a \times \vec b,$$

we get

$$(\vec b+\vec c)\times\vec a=0.$$

Hence $\vec b+\vec c$ is parallel to $\vec a$.

5. So we have:

• $\vec a+\vec c \parallel \vec b$
• $\vec a+\vec b \parallel \vec c$
• $\vec b+\vec c \parallel \vec a$

A simple way to satisfy all three simultaneously is that

$$\vec a+\vec b+\vec c=\vec 0.$$

Let us verify this directly.

If

$$\vec a+\vec b+\vec c=\vec 0,$$

then

$$\vec c=-(\vec a+\vec b).$$

Now,

$$\vec b\times\vec c=\vec b\times[-(\vec a+\vec b)] = -\vec b\times\vec a - \vec b\times\vec b = \vec a\times\vec b.$$

Also,

$$\vec c\times\vec a=[-(\vec a+\vec b)]\times\vec a = -\vec a\times\vec a - \vec b\times\vec a = \vec a\times\vec b.$$

Thus indeed

$$\vec a\times\vec b=\vec b\times\vec c=\vec c\times\vec a.$$

So the required value is

$$\vec a+\vec b+\vec c=\vec 0.$$

6. Therefore, the correct option is:

$$\boxed{\text{C: }0}$$