Fuzzy'ss....

Since $5$ and $6$ are coprime, $\overline{bc}$ is a multiple of $6 \times 5 = 30 \implies c = 0$ and $3 \mid b$

Case 1: $\overline{abc} = \overline{a30}$

According to the rule of divisibility by $7$ , we need $7 \mid \overline{a3}-2 \times 0 = \overline{a3} \implies 7 \mid a-2 \times 3 = a-6$

Since $1 \leq a \leq 9$ , the only possible value is $a = 6 \implies \overline{abc} = 630$

Case 2: $\overline{abc} = \overline{a60}$

$7 \mid a-12 \implies a = 5 \implies \overline{abc} = 560$

Case 3: $\overline{abc} = \overline{a90}$

$7 \mid a-18 \implies a = 4 \implies \overline{abc} = 490$

So there are $\boxed{3}$ fuzzy numbers: $630$ , $560$ and $490$