$\text{ The values of }x\text{ where the function }f\left(x\right)=\frac{\tan x\log (x-2)}{x^2-4x+3}\text{ is discontinuous are given by}$

$f\left(x\right)=\frac{\tan x\log (x-2)}{x^2-4x+3},\text{ being product and quotient of functions }\tan x$

$\log (x-2)\text{ and }\left(x^2-4x+3\right)\text{ must be continuous in its domain of definition. }$

$\tan x\text{ is discontinuous in }\left\{\right(2n+1)\pi /2:n\in Z\}$

$\log (x-2)\text{ discontinuous for }x\le 2$

$\text{ and }{x}^2-4x+3=0\text{ for }x=1\text{ and }3\text{. }$

$\text{ Hence }f\left(x\right)\text{ is discontinuous in }[-\mathrm{\infty },2]\cup \{3,n\pi +\pi /2:n\in Z\}$

$[-\mathrm{\infty },2]\cup \{3,n\pi +\pi /2:n\ge 1\}$

Hence (3) is the correct answer.