jmc

We start with the first equation. Any value of $x$ that makes the first equation true must also satisfy $$(x+a)(x+b)(x+12)=0.$$Therefore, the only possible roots of the first equation are $-a,$ $-b,$ and $-12.$ Because the first equation has three distinct roots, it must be that $-a,$ $-b,$ and $-12$ are all distinct and all satisfy the first equation. This means that $-a,$ $-b,$ and $-12$ cannot equal $-3,$ since when $x=-3$ in the first equation, the denominator of the fraction becomes zero. In conclusion, from the first equation having $3$ distinct roots, we discern that all of the numbers $-a,$ $-b,$ $-12,$ and $-3$ are distinct. That is, all the numbers $a,$ $b,$ $3,$ and $12$ are distinct.

Then $-3$ is necessarily a root of the second equation, because when $x=-3,$ the numerator is zero, while the denominator is nonzero. Thus, $-3$ must be the only root of the second equation. In particular, neither $-2a$ nor $-6$ can be another distinct root of the equation, even though they are roots of the numerator.

Since $-6\ne -3,$ it must be that $-6$ is not a root of the second equation at all, because it makes the denominator zero. Then we must have $-6+b=0,$ so $b=6.$

For $-2a$ not to be another distinct root, we must either have $-2a=-3$ (so that $-2a$ is a root of the second equation, but it is equal to the other root, $-3$), or $x=-2a$ must make the denominator zero. The denominator is $(x+6)(x+12)=0,$ so either $-2a+6=0$ or $-2a+12=0,$ which means either $a=3$ or $a=6.$ But we know that $a,$ $b,$ $3,$ and $12$ are distinct, and $b=6,$ so this is impossible. Hence $-2a=-3,$ so $a=\frac{3}{2}.$

In conclusion, the two equations are $$\frac{(x+\frac{3}{2})(x+6)(x+12)}{(x+3{)}^2}=0$$and $$\frac{(x+3)(x+3)(x+6)}{(x+6)(x+12)}=0,$$which satisfy the conditions: the first equation has roots $x=-\frac{3}{2},-6,-12,$ while the second equation has only the one root $x=-3.$ Hence, $$100a+b=100\left(\frac{3}{2}\right)+6=\boxed{156}.$$