 Solution

Use the orientation induced by the boundary order $O\to A\to B\to O$. The triangle's vector area element integrates to $\frac{1}{2}\mathbf{a}\times \mathbf{b}$. Every point $\mathbf{r}$ in its plane is a linear combination of $\mathbf{a},\mathbf{b}$, so $\mathbf{r}\cdot (\mathbf{a}\times \mathbf{b})=0$. The radial term of the field therefore contributes no flux. The constant term givesReversing the orientation reverses this sign. If $\mathbf{a}$ and $\mathbf{b}$ are collinear, the triangle has zero area and the displayed value is zero by the degenerate-area convention.