Let $b=ra.$ Then your construction gives $$\mu(A_1\cap A_2\cap A_3)=0,\frac{(5r-2)^2}{8r}a \text{ or }(2r-1)a$$according as $r\le\frac25,\frac25\le r \le \frac23$ or $\frac23 \le r.$

Let $b=ra.$ Then your construction gives $$\mu(A_1\cap A_2\cap A_3)=0,\frac{(5r-2)^2}{8r}a \text{ or }(2r-1)a$$according as $r\le\frac25,\frac25\le r \le \frac23$ or $\frac23 \le r.$

Thanks to the nice illustration I can say that as $r$ decreases from $r=\frac12$ to $r=\frac25,$ the small triangle shrinks to a point and beyond that there is an empty triangle in the middle. But as it increases the triangle of overlap grows and touches the sides at $r=\frac23$ and beyond that becomes a hexagon of overlap.