The sum of the ages

$\dfrac{23}{9}=\dfrac{A+B+C+60}{A+B+C-24}$ $\color{#D61F06}(1)$

$B=2+A$ $\color{#D61F06}(2)$

$C=2B$ $\color{#D61F06}(3)$

Simplifying $\color{#D61F06}(1)$ , we get

$\dfrac{23}{9}=\dfrac{A+B+C+60}{A+B+C-24}$

$23A+23B+23C-552=9A+9B+9C+540$

$14A+14B+14C=1092$

Substitute $\color{#D61F06}(2)$ and $\color{#D61F06}(3)$ into the above equation

$14A+14(2+A)+14(2B)=1092$

$14A+14(2+A)+14(2)(2+A)=1092$

$14A+28+14A+56+28A=1092$

$56A=1008$

$A=18$

It follows that $B=20$ and $C=40$