Solution
= Solution
The prescribed <orientation> orders the two edges from $A$ as $\mathbf c-\mathbf a$ followed by $\mathbf b-\mathbf a$. The <vector area> is half their <cross product>:
$$
\mathbf S=\frac12(\mathbf c-\mathbf a)\times(\mathbf b-\mathbf a)
=\frac12(-1,0,1)^T\times(-1,1,0)^T
=\boxed{-\frac12(1,1,1)^T}.
$$
The sign agrees with the <right-hand rule> for the traversal $A\to C\to B\to A$.