Let coordinate of point $A\left(t^2,2t\right) \left(\because a=1\right)$

equation of tangent at point $A$

$yt=x+t^2$

$x-ty+t^2=0$

centre of circle $\left(3,0\right)$

Now Perpendicular distance = radius

$\left|\frac{3-0+t^2}{\sqrt{1+t^2}}\right|=3$

${\left(3+t^2\right)}^2=9\left(1+t^2\right)$

$9+t^4+6t^2=9+9t^2$

$t=0,-\sqrt{3},\sqrt{3}$

So point $A\left(3,2\sqrt{3}\right)$

$\Rightarrow a=3, b=2\sqrt{3}$

(Since it lies in first quadrant)

For point $B$ which is foot of perpendicular from centre $\left(3,0\right)$ to the tangent $x-\sqrt{3}y+3=0$

$\frac{c-3}{1}=\frac{d-0}{-\sqrt{3}}=\frac{-\left(3-0+3\right)}{4}$

$\Rightarrow c=\frac{3}{2} d=\frac{3\sqrt{3}}{2}$

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