$\frac{1}{2}\left(x+1\right)+\frac{1}{4}.\left(x+3\right)=3-\frac{1}{3}\left(x+2\right)$
$\Rightarrow \frac{1}{2}x+\frac{1}{2}+\frac{1}{4}x+\frac{3}{4}=3-\frac{1}{3}x-\frac{2}{3}$
$\Rightarrow \frac{1}{2}x+\frac{1}{4}x+\frac{1}{3}x=3-\frac{2}{3}-\frac{1}{2}-\frac{3}{4}$
$\Rightarrow \frac{13}{12}x=\frac{13}{12}$
=> x = 1