Stacked Fractions

Multiplying the equations, we find $\frac{a^2}{b^2} = \frac{42\cdot 7}{5\cdot 270} = \frac{(2\cdot 3 \cdot 7)\cdot 7}{5 \cdot (2 \cdot 3^3\cdot 5)} = \frac{7^2}{3^2\cdot 5^2};$ $\frac a b = \frac{7}{3\cdot 5} = \frac 7{15}.$ This fraction is reduced to the smallest possible terms, guaranteeing that $a + b = 7 + 15 = 22$ is minimal.

Dividing the equations, we find $c^2 = \frac{42\cdot 270}{5\cdot 7} = \frac{(2\cdot 3\cdot 7)\cdot (2\cdot 3^3\cdot 5)}{5 \cdot 7} = 2^2\cdot 3^4;$ $c = 2\cdot 3^2 = 18.$

Thus, $a + b + c = 7 + 15 + 18 = \boxed{40}$ .

$\begin{cases} \dfrac a{\frac bc} = \dfrac {ca}b = \dfrac {42}5 & ...(1) \\ \dfrac {\frac ab}c = \dfrac a{bc} = \dfrac 7{270} & ...(2) \end{cases}$

$\begin{aligned} \frac {(2)}{(1)}: \quad \frac {ca}b \times \frac {bc}a & = \frac {42}5 \times \frac {270}7 \\ c^2 & = 324 \\ \implies c & = 18 \end{aligned}$

$\begin{aligned} (2): \quad \frac a{18b} & = \frac 7{250} \\ \frac ab & = \frac 7{15} \\ \implies a & = 7 \\ \implies b & = 15 \end{aligned}$

$\implies a+b+c = 7+15+18 = \boxed{40}$