smc

From the Remainder Theorem, we have $2(-2{)}^3-h(-2)+k=0$ and $2(1{)}^3-h(1)+k=0$. Simplifying both of those equations gives $-16+2h+k=0$ and $2-h+k=0$. Since $k=16-2h$ and $k=h-2$, we set those equal to get: $16-2h=h-2$ $3h=18$ $h=6$ This gives $k=4$ when substituting back into either of the two equations, and $|2h-3k|=0$, which is answer $\boxed{0}$.