If  $a,b\in R$  and  $a{x}^2+bx+6=0,a\ne 0$ does not have two distinct real roots, the

Here,  $D\le 0$
and $f\left(x\right)\ge 0,\forall x\in RC$ 
 $\therefore f\left(3\right)\ge 0$  
 $\Rightarrow 9a+3b+6\ge 0$
Or  $3a+b\ge -2$
$\Rightarrow$  Minimum value of 3a+b is -2.
and  $f\left(6\right)\ge 0$
 $\Rightarrow 36a+6b+6\ge 0$
 $\Rightarrow 6a+b\ge -1$
$\Rightarrow$  Minimum value of  $6a+b$  is -1