You Can't Probably Just Do Trial and Error

We note that

$\begin{aligned} a & \equiv 0 \text{ (mod 5)} \\ a & \equiv 0 \text{ (mod 7)} \\ \implies a & \equiv 35x & \small \color{#3D99F6} \text{where }x \text{ is an integer.} \\ a - 1 & \equiv 0 \text{ (mod 3)} \\ a & \equiv 1 \text{ (mod 3)} \\ 35x & \equiv 1 \text{ (mod 3)} \\ \implies x & = 2 \\ \implies a & = 35(2) = 70 \end{aligned}$

Similarly,

$\begin{aligned} b & \equiv 21y & \small \color{#3D99F6} \text{where }y \text{ is an integer.} \\ b & \equiv 1 \text{ (mod 5)} \\ 21y & \equiv 1 \text{ (mod 5)} \\ \implies y & = 1 \\ \implies b & = 21 \end{aligned}$

And

$\begin{aligned} c & \equiv 15z & \small \color{#3D99F6} \text{where }z \text{ is an integer.} \\ c & \equiv 1 \text{ (mod 7)} \\ 15z & \equiv 1 \text{ (mod 7)} \\ \implies z & = 1 \\ \implies c & = 15 \end{aligned}$

$\implies a+b+c = 70+21+15 = \boxed{106}$

Note that:

$\displaystyle \begin{aligned} & (a-1) \mod 3 \equiv 0 \\ & b \mod 3 \equiv 0 \\ & c \mod 3 \equiv 0 \\ \\ & a \mod 5 \equiv 0 \\ & (b-1) \mod 5 \equiv 0 \\ & c \mod 5 \equiv 0 \\ \\ & a \mod 7 \equiv 0 \\ & b \mod 7 \equiv 0 \\ & (c - 1) \mod 7 \equiv 0 \end{aligned}$

First, solve for a:

$\displaystyle \begin{aligned} & a \mod 3 \equiv 1 \Longleftrightarrow a = 3k + 1 \\ & a \mod 5 \equiv 0 \Longleftrightarrow a = 5c \\ & \implies 5c - 3k = 1 \implies c = 2 , k = 3 \\ & c = 2 + 3p \ , \ k = 3 + 5p \\ & \implies a = 15p + 10 \\ \\ & a \mod 7 \equiv 0 \implies a = 7h \\ & \implies 7h = 15p + 10 \implies h = 10 \ , \ p = 4 \\ & \text{smallest a} \ = 7(10) = 70 \end{aligned}$

Then, solve for b:

$\displaystyle \begin{aligned} & b \mod 5 \equiv 1 \Longleftrightarrow b = 5s + 1 \\ & b \mod 3 \equiv 0 \implies b = 3a \\ & \implies 3a = 5s + 1 \implies a = 2 , s = 1 \\ & a = 2 + 5t \ , \ s = 1 + 3t \\ & \implies b = 6 + 15t \\ \\ & b \mod 7 \equiv 0 \implies b = 7y \\ & \implies 7y = 6 + 15t \implies y = 3 \ , \ t = 1 \\ & \text{smallest b} \ = 7(3) = 21 \end{aligned}$

Lastly, solve for c:

$\displaystyle \begin{aligned} & c \mod 7 \equiv 1 \implies c = 7g + 1 \\ & c \mod 3 \equiv 0 \implies c = 3f \\ & \implies 7g + 1 = 3f \implies g = 2 \ , \ f = 5 \\ & \implies g = 2 + 3q \ , \ f = 5 + 7q \\ & \implies c = 15+21q \\ \\ & c \mod 5 \equiv 0 \implies c = 5v \\ & \implies 5v = 15+21q \implies q = 0 \ , \ v = 3 \\ & \text{smallest c} \ = 15 \\ & \therefore \boxed{a + b + c = 70+21+15 = 106} \end{aligned}$