Mixed product of second-order tensors and vectors

Question

I was studying the angular momentum equation in the continuum case and I encountered this identity. I am not sure how the identity is derived. Could some one supply more details and intermediate step? $$\int_s -\vec{n}\cdot(\underline{\sigma}\times\vec{r})dS=\int_s -\vec{n}\cdot\underline{\sigma}\times\vec{r}dS = \int_s \vec{r}\times\vec{n}\cdot\underline{\sigma}dS$$ where $\vec{n}$ and $\vec{r}$ are first-order tensors (vectors) and $\underline{\sigma}$ is a second-order tensor.

Answer 1

I finally found an answer. Basically, there is no order in the operation, thus one can see the dot product as a single object. The anticommutative property odds. Calling $\vec{n}\cdot\underline{\sigma}=\vec{a}$ then: $$ -\vec{a}\times\vec{r}=\vec{r}\times\vec{a}$$