Taking $a = b = c$ we obtain
$$ra^2 + (3 - r) \frac{1}{a} \ge 3 \iff (a - 1)(r(a^2 + a + 1) - 3) \ge 0$$
for any $a > 0$. Then it easily follows that $r = 1$.

Conversely, let $r = 1$. Then we write the inequality as $\frac{2 + abc}{3} \ge \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$. The GM-HM inequality implies that the right hand side does not exceed $\sqrt[3]{abc}$ and, setting $x = \sqrt[3]{abc} > 0$, it is enough to prove that
$$\frac{2+x^3}{3} \ge x$$
which is equivalent to the obvious $(x-1)^2(x+2) \ge 0$.