If $\left(\alpha ,\beta \right),\left(\beta ,\gamma \right)$  and $\left(\gamma ,\alpha \right)$  are respectively the roots of $x^2-2px+2=0,x^2-2qx+3=0$  and $x^2-2rx+6=0$ , where $\alpha ,\beta ,\gamma$  are all positive, then the value of $p+q+r$  is ______

$\alpha +\beta =2p,\alpha \beta =2$

$x^2-2qx+3=0$  roots $\beta ,\gamma$

$\beta +\gamma =2q,\beta \gamma =3$

$x^2-2rx+6=0$  roots $\gamma ,\alpha$

$\gamma +\alpha =2r,\gamma \alpha =6$

$\Rightarrow p+q+r=\frac{\alpha +\beta }{2}+\frac{\beta +\gamma }{2}+\frac{\gamma +\alpha }{2}=\alpha +\beta +\gamma$

$\Rightarrow \gamma \alpha =6,\alpha \beta =2,\beta \gamma =3$

$2\times 3=6$

$\therefore \left(\alpha \beta \right)\left(\beta \gamma \right)=\gamma \alpha$

$\alpha {\beta }^2\gamma =\gamma \alpha$

${\beta }^2=1\Rightarrow \beta =\pm 1$

$\alpha ,\beta ,\gamma$  are all positive

$\Rightarrow \beta =1$

$\therefore \alpha =2,\gamma =3$

$\therefore p+q+r=\alpha +\beta +\gamma$

$=1+2+3=6$