$\vec{a} \cdot [(\vec{b} + \vec{c}) \times (\vec{a} + \vec{b} + \vec{c})]$ equals

Use distributivity of the cross product: $$ (\vec{b}+\vec{c})\times(\vec{a}+\vec{b}+\vec{c})=(\vec{b}+\vec{c})\times\vec{a}+(\vec{b}+\vec{c})\times(\vec{b}+\vec{c}) $$ Since the cross product of a vector with itself is zero, $$ (\vec{b}+\vec{c})\times(\vec{b}+\vec{c})=\vec{0} $$ So the expression becomes $$ \vec{a}\cdot[(\vec{b}+\vec{c})\times\vec{a}] $$ Now $[(\vec{b}+\vec{c})\times\vec{a}]$ is perpendicular to $\vec{a}$, hence their dot product is $$ 0 $$ Checking the options, this matches option $\text{A}$.