Step 1: Choosing a different pair of sides:
Instead of \(\overrightarrow{AB},\overrightarrow{AC}\), use Area \(=\dfrac12|\overrightarrow{BA}\times\overrightarrow{BC}|\) — the same triangle, computed from vertex \(B\), which must give an equal magnitude.
Step 2: Substituting:
\(\overrightarrow{BA}=\vec a-\vec b\), \(\overrightarrow{BC}=\vec c-\vec b\), so \((\vec a-\vec b)\times(\vec c-\vec b)=\vec a\times\vec c-\vec a\times\vec b-\vec b\times\vec c+\vec b\times\vec b\).
Step 3: Simplifying:
\(\vec b\times\vec b=\vec 0\); \(-\vec a\times\vec b=\vec b\times\vec a\)... to match the target expression, rewrite: \(\vec a\times\vec c-\vec a\times\vec b-\vec b\times\vec c=-(\vec c\times\vec a)-(\vec a\times\vec b)-(\vec b\times\vec c)=-(\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a)\).
Step 4: Taking magnitudes:
Since magnitude is unaffected by an overall sign flip, \(|\overrightarrow{BA}\times\overrightarrow{BC}|=|\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a|\), matching the vertex-\(A\) computation exactly.
Final Answer:
\[ \boxed{\text{Area}=\dfrac12|\vec a\times\vec b+\vec b\times\vec c+\vec c\times\vec a|} \]