Answer: $(2\cdot 10^k-1,2\cdot 10^k-1,2\cdot 10^k),(5\cdot 10^k,5\cdot 10^k,5\cdot 10^k-1)$ for all integers $0\leq k$: Observation 1: Notice that $c\neq a,b$. Observation 2: Notice that $x!_0<(x+2)!_0$ as only one of $x+1,x+2$ can be divisible by $5$ and the corresponding power of $2$ is smaller then the other. Similarly, for all integers $1<c$ we have $x!_0<(x+c)!_0$. Observation 3: Let $\pi(x)$ denote the largest divisor of $x!$ that is not divisible $2$ or $5$. Notice $\pi(x)$ is non decreasing, $\pi(x)<\pi(x+2)$ (Catalan), $\pi(x)|x!_0$, and $\pi(x)|\pi(y)$ for all $x\leq y$. Noticing that two of $a!_0$, $b!_0$, $c!_0$ are divisible by the median of $\pi(a),\pi(b),\pi(c)$ we must have that the two smallest values amongst $\pi(a),\pi(b),\pi(c)$ are equal. Thus the smallest two out of $a,b,c$ differ by at most one. Notice that if $\pi(x)=\pi(x+1)$ then $5|x+1$. Case 1: $c$ is the largest out of $a,b,c$ If $a$ and $b$ differ by one then the larger must be a multiple of $5$. Then $c!_0$ will be at least twice as large as $a!_0$, $b!_0$ so we must have $a=b$. A simple bounding argument shows that $c-1=a=b$ leading to the above solution. Case 2: $c$ is not the largest of $a,b,c$ If $c$ is in the middle then we get that the three numbers are consecutive. Thus $c$ must be a multiple of $5$, a contradiction. Otherwise $c$ is the smallest and we must have $c+1=a=b$ leading to the above solution. *will add in more details later