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

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

$\text{ The image of point }(a,b)\text{ in }y\text{-axis is }(-a,b)$

$\text{ Therefore, the image of }(1,1)\text{ in }y\text{-axis is }(h,k)=(-1,1)$

$\text{ It implies that }h+k=0$