Inverse of Root

$$ The inverse function of $y=\sqrt{x-3}+6$ can be written as $$ \begin{aligned} y&=\sqrt{x-3}+6\quad\quad\quad\tag1\\ y-6&=\sqrt{x-3}\\ (y-6)^2&=x-3\\ y^2-12y+36&=x-3\\ x&=y^2-12y+39\\ f^{-1}(x)&=x^2-12x+39\quad\quad\quad\tag2 \end{aligned} $$ Based on the last equation, we can obtain that $a=-12$ and $b=39$ . To obtain value of $c$ , take a look equation $(1)$ . The minimum value of $x$ such that $y\in\mathbb{R}$ is equal to $3$ , therefore the minimum value of $y$ is equal to $6$ . Since in inverse function $y$ becomes $x$ (see equation $(2)$ ), then, from the given condition $x\ge c$ , the minimum value of $c$ is equal to $6$ . Thus, the possible value of $a+b+c$ based on the given options is $-12+39+6=\boxed{33}$ . $$

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$