$\because \left(x-1\right)^{2}+|y+2|=0$，
$\therefore x-1=0$，$y+2=0$，
解得$x=1$，$y=-2$，
$\because z$是最小的正整数，
$\therefore z=1$，
$\therefore x+2y+3z=1+2\times \left(-2\right)+3\times )=1-4+3=0$，
故答案为：$0$.

$\because \left(x-1\right)^{2}+|y+2|=0$，
$\therefore x-1=0$，$y+2=0$，
解得$x=1$，$y=-2$，
$\because z$是最小的正整数，
$\therefore z=1$，
$\therefore x+2y+3z=1+2\times \left(-2\right)+3\times )=1-4+3=0$，
故答案为：$0$.