If the function $f\left(x\right)=x^3-6x^2+ax+b$  defined on $[1,3]$  satisfies the rolle’s theorem for $c=\frac{(2\sqrt{3}+1)}{\sqrt{3}}$ , the :

$f\left(x\right)=x^3-6x^2+ax+b$  satisfies rolle’s theorem for $c=\frac{(2\sqrt{3}+1)}{\sqrt{3}}$  in $[1,3]$

$\therefore f\left(1\right)=f\left(3\right)\Rightarrow a=11$  and $f\text{'}\left(c\right)=0\Rightarrow 3c^2-12c+a=0$ , which is true for $c=\frac{(2\sqrt{3}+1)}{\sqrt{3}}$  and every real value of $b$ .

Hence, $a=11,b\in R$ .