We let $x = AP = s - a, y = BQ = s-b, z = CR = s-c$, and without loss of generality let $x \le y \le z$. Then $x + y + z = 3s - (a+b+c) = s$, so $r = \frac {A}{s} = \frac{\sqrt{(x+y+z)xyz}}{x+y+z} = \sqrt{\frac{xyz}{x+y+z}}$. Thus,
$$6\sqrt{\frac{x+y+z}{xyz}} = \frac{2yz + 5xy + 5xz}{xyz}$$
Squaring and simplifying yields (after much grueling work)
$$0 = 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz$$
We claim that the inequality
\begin{align}0 \le 4y^2z^2 + 25x^2y^2 + 25x^2z^2 - 16xy^2z - 16xyz^2 + 14x^2yz \tag{*} \end{align}
holds true, with equality iff $4x = y = z$.

Note that $(*)$ is homogeneous in $x,y,z$, so without loss of generality, scale so that $x=1$. Then
$$0 \le y^2 (4z^2 - 16z + 25) + y(14z - 16z^2) + 25z^2$$
which is a quadratic in $y$. As $4z^2 - 16z + 25 = 4(z-4)^2 + 9 \ge 0$, it suffices to show that the quadratic cannot have more than one root, or the discriminant $\Delta \le 0$. Then,
$$\Delta = z^2(14-16z)^2 - 4(4z^2 - 16z + 25)(25z^2) = -144(z-4)^2 \le 0$$
as desired. Equality comes when $z = 4$; since $(*)$ is symmetric in $y$ and $z$, it follows that $y = 4$ is also necessary for equality. Reversing our scaling, it follows that $x:y:z = 1:4:4$.
Then $s = x+y+z = 9x$, and $s-a = x, s-b = 4x, s-c = 4x$ yields $a:b:c = 8:5:5$. Thus, we have proved that all possible ABC are isosceles (as b = c), and that they are similar to a 5-5-8 triangle.
$\blacksquare$