Simultaneous multiplying

Answer: $10\frac12$

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)

$b^2:\quad ab\times bc = b^2\times ca = b^2\times 3\qquad \therefore \ b^2=\frac{2\times24}{3}=16\quad \Rightarrow b=4$

$c^2:\quad c^2=\frac{3\times24}{2}=\frac36\quad \Rightarrow c=6$

$a+b+c=\frac12+4+6=10\frac12$

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 $a+b+c=\frac{1}{2}+4+6=10\frac{1}{2}$.

Solving by substitution 

$\ \ ab=2\\

\Rightarrow a=\tfrac2b \qquad\qquad ca =3 \Rightarrow c\times \tfrac 2b =3 \\

\qquad\qquad\ \ \ \qquad \qquad\qquad\Rightarrow 2c=3b \\

\qquad\ \ \ \qquad\qquad \qquad \qquad\Rightarrow c=\tfrac32b\qquad \qquad bc=24 \Rightarrow \tfrac32b^2=24\\

\qquad\qquad\ \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad\Rightarrow b^2=36\\

\qquad\qquad\ \qquad\qquad\qquad\qquad\qquad\qquad \qquad\qquad \qquad\Rightarrow b=6$