The number is
because we can rewrite the three given conditions as follows, where $H$, $T$, and $U$ are the hundreds, tens, and units digits respectively:
$H+U-T=8,\quad 3H+2T-U=33,\quad 100H+10T+U=53H+53T+53U$
Adding the first two gives
$4H+T=41$, therefore $T=41-4H$. Since both $T$ and $H$ must be numbers from $0$ to $9$, the only possibilities are $H=8,T=9$ and $H=9,T=5$.
Also from the first equation,
$U=8+T-H=8+(41-4H)-H=49-5H$. Here the only possibilities are $H=8,U=9$ and $H=9,U=4$.
So the number must be
either $899$ or $954$. Only one of these is a multiple of $53$, namely $954=53\times18=53\times(9+5+4)$, so the problem is solved.