Let $\alpha =\frac{a}{r}, \beta =a$ and $\gamma =ar$ are the roots of $3x^3-26x^2+52x-24=0$

Thus products of roots 

$a^3=\frac{24}{3}=8$

$\Rightarrow a=2$

And sum of roots 

$\frac{2}{r}+2+2r=\frac{26}{3}$

$\Rightarrow \frac{1}{r}+r=\frac{10}{3}$

$\Rightarrow 3r^2-10r+3=0$

$\Rightarrow \left(3r-1\right)\left(r-3\right)=0$

$\Rightarrow r=\frac{1}{3}, 3$

But $\left|r\right|<1$,so $r=\frac{1}{3}$

Hence, the roots are

$\alpha =\frac{2}{3}, \beta =2, \gamma =6$

Therefore,

$3\alpha +2\beta +\gamma =2+4+6=12$