The image of centroid of the triangle formed by the points  $\left(0,0\right),\left(1,2\right),\left(2,1\right)$ in  y- axis is $\left(h,k\right)$   then $h+k$ is

The centroid of the triangle is G

$\begin{array}{c}G=\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)\\ =\left(\frac{0+1+2}{3},\frac{0+2+1}{3}\right)\\ =\left(1,1\right)\end{array}$

The image of point $\left(a,b\right)$  in y- axis is  $\left(-a,b\right)$
Therefore, the image of  $\left(1,1\right)$ in y- axis is  $\left(\mathrm{h},\mathrm{k}\right)=\left(-1,1\right)$
It implies that  $\mathrm{h}+\mathrm{k}=\boxed{0}$