We have,

$y^2=6x$

Comparing with $y^2=4ax$, we get $a=\frac{3}{2}$

Let the point on parabola be $P\equiv \left(a{t}^2,2at\right)\equiv \left(\frac{3}{2}{t}^2,3t\right)$.

Normal at point $P$ is

$tx+y=3t+\frac{3}{2}{t}^3$

It passes through $\left(3,\frac{3}{2}\right)$, hence

$\Rightarrow 3t+\frac{3}{2}=3t+\frac{3}{2}{t}^3$

$\Rightarrow t^3=1\Rightarrow t=1$

Hence,

$P\equiv \left(\frac{3}{2},3\right)=(\alpha ,\beta )$

Therefore,

$2(\alpha +\beta )=2\left(\frac{3}{2}+3\right)=9$