Diophantine's Story of L.C.M.

Let $gcd(a,b,c)=g$

$\therefore$ let $a=gx,b=gy,c=gz$

Then, we have $lcm(a,b,c)=gxyz$

$\Rightarrow gx+gy+gz=gxyz$

$\Rightarrow x+y+z=xyz$

By symmetry, without a loss in generality, we assume that $x\leq y\leq z$

$\therefore x+y+z\leq z+z+z=3z$

$\because x+y+z=xyz$

$\Rightarrow xyz\leq 3z$

$\Rightarrow xy\leq3$

The possible soutions for $x,y$ are $x=1,y=1$ and $x=1,y=2$ and $x=1,y=3$

When $x=1,y=1$ ,

$1+1+z=z$ , which is impossible to hold true.

$\therefore$ $x=1,y=1$ is unsuitable.

When $x=1,y=2$ ,

$1+2+z=2z$

$\Rightarrow z=3$

When $x=1,y=3$ ,

$1+3+z=3z$

$2z=4$

$z=2$

We conclude that all the possible permutations of $1,2,3$ are the solutions for $x,y,z$ when the restriction of $x\leq y\leq z$ is lifted off.

$\therefore a=g, b=2g, c=3g$ (Not to mention, the other permutations are included too.)

When $a=12,$

$g=12$

$b+c=2g+3g=5g=5(12)=60$

When $b=12$ ,

$2g=12$

$g=6$

$a+c=g+3g=4g=4(6)=24$

When $c=12$ ,

$3g=12$

$g=4$

$b+c=g+2g=3g=3(4)=12$

Coming back to our question,

The sum of all possible $a+b=60+24+12=\boxed{96}$