Middle number is $n$

$\begin{align}n-2\ \ +\ \ n\ \ +\ \ n+12 &= 100\\

\Rightarrow 3n+10&=100\\

\Rightarrow 3n&=90\\

\Rightarrow n&=30\end{align}$

So $n=30$, $n+12=42$ and $n-2=28$.

Try some numbers which are the right distances apart

$\begin{align}30 + 32 + 44 &= 106\\

&=100 + 6\\

&= 100+3\times2\end{align}$

$\therefore (30-2)+(32-2)+(44-2)$ should work

$28+30+42=100$

Try some numbers which add up to $100$

$33+33+34=100$

Need to make largest number larger, always balancing out changes:

$(33-2)+\ \quad33\ \quad+(34+2)=31+33+36=100$

$(31-1)+(33-1)+(36+2)=30+32+38=100$

$(30-2)+(32-2)+(38+4)=28+30+42=100$

You can find more short problems, arranged by curriculum topic, in our short problems collection.