解:设$AF = x.$
$\because \triangle ABC$的内切圆$\odot O$与三边分别相切于$D,$$E,$$F$三点,$AB = 7,$$BC = 12,$$CA = 11,$
$\therefore AE = AF = x,$$BF = BD = AB - AF = 7 - x,$$CE = CD = AC - AE = 11 - x.$
$\because BD + CD = BC,$
$\therefore 7 - x + 11 - x = 12,$
$18 - 2x = 12,$
$-2x = 12 - 18,$
$-2x = -6,$
解得$x = 3.$
$\therefore AF = 3,$$BD = 7 - x = 4,$$CE = 11 - x = 8$