If $f\left(x\right)$  be continuous function for all real values of $x$  and satisfies; $x^2+\left(f\left(x\right)-2\right)x+2\sqrt{3}-3-\sqrt{3} f\left(x\right)=0,$ $\forall x\in R.$  Then the value of $f\left(\sqrt{3}\right)$  is

A s$f\left(x\right)$ is continuous for all $x\in R$ 

Thus, $\underset{x\to \sqrt{3}}{\lim }f\left(x\right)=f\left(\sqrt{3}\right)$

Where $f\left(x\right)=\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x},x\ne \sqrt{3}$

$\underset{x\to \sqrt{3}}{\lim }f\left(x\right)=\underset{x\to \sqrt{3}}{\lim }\frac{x^2-2x+2\sqrt{3}-3}{\sqrt{3}-x}$

$=\underset{x\to \sqrt{3}}{\lim }\frac{\left(2-\sqrt{3}-x\right)\left(\sqrt{3}-x\right)}{\left(\sqrt{3}-x\right)}=2\left(1-\sqrt{3}\right)$

$\Rightarrow f\left(\sqrt{3}\right)=2\left(1-\sqrt{3}\right)$