Solving by elimination
$a^2:\quad ab\times ca = a^2\times bc = a^2\times 24\qquad \therefore \ a^2=\frac{2\times3}{24}=\frac14\quad \Rightarrow a=\frac12$ (all are positive)
$b^2:\quad ab\times bc = b^2\times ca = b^2\times 3\qquad \therefore \ b^2=\frac{2\times24}{3}=16\quad \Rightarrow b=4$
Multiplying all of the equations
$ab\times bc\times ca = (abc)^2=2\times 24\times 3=144$
$abc$ is positive so $abc=12$
Then the third equation tells us that $b=4$, the second that $a=\frac{1}{2}$ and the first that $c=6$. Therefore $a+b+c=\frac{1}{2}+4+6=10\frac{1}{2}$.