We first expand the left-hand-side of the third equation to get $(a+b)(a+c)=a^{2}+ac+ab+bc=24$. From this, we subtract the second equation to obtain $a^{2}=4$, so $a=\pm 2$. If $a=2$, plugging into the first equation gives us $b+c=98$ and plugging into the second equation gives us $2(b+c)+bc=20 \Rightarrow 2(98)+bc=20 \Rightarrow bc=-176$. Then, if $a=-2$, plugging into the first equation gives us $b+c=102$, and plugging into the second equation gives us $-2(b+c)+bc=20 \Rightarrow -2(102)+bc=20 \Rightarrow bc=224$. Therefore, the possible values of $bc$ are $224,-176$.