If the locus of $ z \in \mathbb{C} $, such that
$
\text{Re} \left( \frac{z - 1}{2z + i} \right) + \text{Re} \left( \frac{ \bar{z} - 1}{2 \bar{z} - i} \right) = 2,
$
is a circle of radius $ r $ and center $ (a, b) $, then
$
\frac{15ab}{r^2} \text{ is equal to:}
$