smc

Let $a$, $b$, and $c$ be the side lengths of the triangle such that $a<b<c$. We are given $a+b>c$ by the triangle inequality. Let $h_a$, $h_b$, and $h_c$ be the altitudes to sides $a$, $b$, and $c$ respectively. We see that $h_c<h_b<h_a$. By computing the areas using $a$, $b$, and $c$ as bases we get $$\frac{1}{2}a{h}_a=\frac{1}{2}b{h}_b=\frac{1}{2}c{h}_c$$Solving for $a$ and $b$, plugging back into the triangle inequality, and canceling $c$ from both sides leaves us with $$\frac{h_c}{h_a}+\frac{h_c}{h_b}>1$$ Further manipulation gives $$h_c>\frac{h_a{h}_b}{h_a+h_b}$$ Looking at the answer choices and letting $h_c$ be the smallest value each time, we see that $5>\frac{12\cdot 13}{12+13}$ is not true. Thus, the answer is $\boxed{5,12,13}$.