jmc

Substituting in for $d$ in terms of $c$ in the second equation gives $b=3(2c-1)=6c-3$. Substituting in for $b$ in terms of $c$ in the third equation gives $a=3(6c-3)+c=19c-9$. Finally, substituting in for $a$, $b$, and $d$ in terms of $c$ in the first equation gives $2(19c-9)+3(6c-3)+5c+7(c-1)=34$. Simplifying this gives $68c=68$, so $c=1$. Note that $c-1=d$, so $d=0$. Therefore, the product $a\cdot b\cdot c\cdot d=\boxed{0}$.