Trio Fighting

Since we are looking for unordered triples, we might as well assume that $0 \le a \le b \le c$ . Then $100 = ab + ac + bc \ge 3a^2$ , so that $0 \le a \le 5$ .

If $a=0$ we want $bc = 100$ , and we obtain $5$ solutions: $(0,1,100)$ , $(0,2,50)$ , $(0,4,25)$ , $(0,5,20)$ , $(0,10,10)$ .

If $a=1$ we want $(b+1)(c+1) = 101$ . This equation has no solutions with $c \ge b \ge 1$ .

If $a=2$ we want $(b+2)(c+2) = 104 = 2^3 \times 13$ , and we obtain $2$ more solutions: $(2,2,24)$ , $(2,6,11)$ .

If $a=3$ we want $(b+3)(c+3) = 109$ . This equation has no solutions with $c \ge b \ge 3$ .

If $a=4$ we want $(b+4)(c+4) = 116 = 2^2 \times 29$ . This equation has no solutions with $c \ge b \ge 4$ .

If $a=5$ we want $(b+5)(c+5) = 125 = 5^3$ . This equation has no solutions with $c \ge b \ge 5$ .

Thus there are exactly $\boxed{7}$ unordered solutions.