abc ... not d?

$\overline { aaaaaa } can\quad be\quad written\quad as\quad a(111111)$ . Factoring 111111 gives us
$111111=3\times 7\times 11\times 13\times 37$ $\Rightarrow b\times c\times \overline { bc } \times (2c-b)\times (2b+c)\quad has\quad the\quad factors\quad 3,7,11,13,37\\ \Rightarrow b\times c\times (10b+c)\times (2c-b)\times (2b+c)\quad has\quad the\quad factors\quad 3,7,11,13,37\\ \quad$
Since b and c are single digit numbers, only 10b+c is a multiple of 37
$\Rightarrow 10b+c=74\quad or\quad 37 \quad \quad If\quad 74,then,\quad b=7,c=4\\ \Rightarrow b\times c\times (10b+c)\times (2c-b)\times (2b+c)=7\times 4\times 74\times 1\times 18$

But nor 11 neither 13 are its factors. $If\quad 37,\quad then,b=3,c=7\\ \Rightarrow b\times c\times (10b+c)\times (2c-b)\times (2b+c)=3\times 7\times 37\times 11\times 13=111111$ Hence a=1, b=3, c=7. $\Rightarrow abc=\boxed { 21 }$