If $\vec{u},\vec{v}$ and $\vec{w}$ are three non-coplanar vectors, then $(\vec{u}+\vec{v}-\vec{w})\cdot [(\vec{u}-\vec{v})\times (\vec{v}-\vec{w})]$ equals:

Let the expression be $E$.
Inside the bracket: $(\vec{u}-\vec{v})\times (\vec{v}-\vec{w})=\vec{u}\times \vec{v}-\vec{u}\times \vec{w}-\vec{v}\times \vec{v}+\vec{v}\times \vec{w}$
Since $\vec{v}\times \vec{v}=0$:$=\vec{u}\times \vec{v}+\vec{w}\times \vec{u}+\vec{v}\times \vec{w}$
Now, dot this with $(\vec{u}+\vec{v}-\vec{w})$:$E=\vec{u}\cdot (\vec{v}\times \vec{w})+\vec{v}\cdot (\vec{w}\times \vec{u})-\vec{w}\cdot (\vec{u}\times \vec{v})$
Note: Other terms like $\vec{u}\cdot (\vec{u}\times \vec{v})$ are zero because of orthogonal vectors in scalar triple product.
Since $[\vec{u}\vec{v}\vec{w}]=[\vec{v}\vec{w}\vec{u}]=[\vec{w}\vec{u}\vec{v}]$:$E=[\vec{u}\vec{v}\vec{w}]+[\vec{u}\vec{v}\vec{w}]-[\vec{u}\vec{v}\vec{w}]=[\vec{u}\vec{v}\vec{w}]=\vec{u}\cdot (\vec{v}\times \vec{w})$.