1. **State the problem:** Find the vector sum $\vec{AB} + \vec{BC} + \vec{CA}$ where $A$, $B$, and $C$ are vertices of a triangle. 2. **Recall vector definitions:** The vector from point $P$ to point $Q$ is $\vec{PQ} = \vec{OQ} - \vec{OP}$, where $\vec{OP}$ and $\vec{OQ}$ are position vectors of points $P$ and $Q$ respectively. 3. **Express each vector:** $$\vec{AB} = \vec{B} - \vec{A}$$ $$\vec{BC} = \vec{C} - \vec{B}$$ $$\vec{CA} = \vec{A} - \vec{C}$$ 4. **Sum the vectors:** $$\vec{AB} + \vec{BC} + \vec{CA} = (\vec{B} - \vec{A}) + (\vec{C} - \vec{B}) + (\vec{A} - \vec{C})$$ 5. **Simplify by grouping terms:** $$= \vec{B} - \vec{A} + \vec{C} - \vec{B} + \vec{A} - \vec{C}$$ 6. **Cancel terms:** $$= \cancel{\vec{B}} - \vec{A} + \vec{C} - \cancel{\vec{B}} + \vec{A} - \vec{C}$$ $$= (- \vec{A} + \vec{A}) + (\vec{C} - \vec{C})$$ $$= \vec{0} + \vec{0} = \vec{0}$$ 7. **Conclusion:** The sum of the vectors around the triangle is the zero vector: $$\boxed{\vec{AB} + \vec{BC} + \vec{CA} = \vec{0}}$$ This result makes sense because moving along the triangle edges in order returns you to the starting point, so the net displacement is zero.