Too Many Solutions!

given a+b+c+d = 12 ............ (1)
and abcd=27+ab+ac+ad+bc+bd+cd ................ (2)
applying AM-GM inequality $\frac{ab+ac+ad+bc+bd+cd}{6}$ >= $(abcd)^{1/2}$ ;
ab+ac+ad+bc+bd+cd>= 6 $\sqrt{abcd}$
abcd >= 27+ 6 $\sqrt{abcd}$ (using (2))
abcd- 6 $\sqrt{abcd}$ -27>=0
( 6 $\sqrt{abcd}$ -9)( 6 $\sqrt{abcd}$ +3) >=0
$\sqrt{abcd}$ >=9 .............(3)
$\frac{a+b+c+d}{4}$ >= $(abcd)^{1/4}$ a+b+c+d=12 .........(1)
9>= $\sqrt{abcd}$ .................(4)
from (3) and (4) we get abcd=81. thus AM=GM
thus a=b=c=d=3 since there is only one such solution, the answer is 3.

$12=a+b+c+d \ge 4(abcd)^\frac{1}{4} \Rightarrow abcd \le 81$ $abcd=27+ab+ac+ad+bc+bd+cd \ge 27+6(abcd)^\frac{1}{2}$ $\Rightarrow abcd \Rightarrow 81\ge abcd=81\Rightarrow a=\boxed{3}$