Solution
= Solution
The two edge <vectors> based at $A$ are $\mathbf b-\mathbf a=(-3,-1,3)$ and $\mathbf c-\mathbf a=(-3,2,2)$. Their <cross product> is $(-8,-3,-9)$, with squared <Euclidean norm> $64+9+81=154$. The <triangle> occupies half the corresponding <parallelogram>, so
$$
\boxed{\text{area}=\frac12| (\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)|=\frac{\sqrt{154}}2.}
$$