Let S be the set of all real roots of the equation

Solution:

Let $3^{x}=y$

$\therefore \quad y(y-1)+2=|y-1|+|y-2|$

Case 1: when $y>2$

$y^{2}-y+2=y-1+y-2$

$y^{2}-3 y+5=0$

$\because \quad D<0[\therefore$ Equation not satisfy. $]$

Case 2: when $1 \leq y \leq 2$

$y^{2}-y^{2}+2=y-1-y+2$

$y^{2}-y+1=0$

$\because \quad D<0[\therefore$ Equation not satisfy.]

Case 3: when $y \leq 1$

$y^{2}-y+2=-y+1-y+2$

$y^{2}+y-1=0$

$\therefore y=\frac{-1+\sqrt{5}}{2}$

$=\frac{-1-\sqrt{5}}{2}$ $[\therefore$ Equation not Satisfy $]$

$\therefore$ Only one $-1+\frac{\sqrt{5}}{2}$ satisfy equation