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Multiply the first equation by $2$ throughout we have:

$10a+10b+4ab = 184 \quad \Rightarrow (2a+5)(2b+5)=209=(11)(19)$

Since $a$ and $b$ are integers, it means:

$(2a)(2b)=(11-5)(19-5)=(6)(14)\quad \Rightarrow ab=(3)(7)$

Similarly,

$10b+10c+4bc = 272 \quad \Rightarrow (2b+5)(2c+5)=297=(11)(27)$

$\Rightarrow bc=(3)(11)$

And

$10c+10a+4ca = 488 \quad \Rightarrow (2b+5)(2c+5)=513=(19)(27)$

$\Rightarrow ca=(7)(11)$

As both $b$ and $3$ appear in Eqn 1 and Eqn 2, $b=\boxed{3}$

Subtracting pairs of equations, we get:

(c - a) * (2b + 5) = 44 or, c - a = 44 / (2b + 5)

(a - b) * (2c + 5) = 108 or, a - b = 108 / (2c + 5)

(b - c) * (2a + 5) = -152 or, b - c = -152 / (2a + 5)

Note that each of these 3 quantities is an integer as they are the differences of 2 integers, which must be integers. So, (2b + 5 ) | 44, (2c + 5) | 108, and (2a + 5) | 152

Since a > 0, b > 0, c > 0, we have 2a + 5 > 5, 2b + 5 > 5, 2c + 5 > 5

Now, 44 has factors 1, 2, 4, 11, 22 and 44. So, 2b + 5 can be equal to 11, 22 or 44. A little trial will yield the solution b = 3, a = 7, and c = 11