The midpoint of the chord $x+y=2$  with respect to the ellipse $3x^2+2y^2=6$ is

Let  $x+y=\mathrm{2..............}\left(1\right)$
$P\left(x_1{y}_1\right)$  is a midpoint then equation of chord 
$3x{x}_1+2y{y}_1=3x_1^2+2y_1^2\mathrm{.............}\left(2\right)$ 
Comparing (1) & (2) 

$$\begin{gathered} \frac{3x_1}{1}=\frac{2y_1}{1}=\frac{3{x^2}_1+2{y^2}_1}{2} \\ \Rightarrow \frac{3x_1}{1}=\frac{2y_1}{1}\Rightarrow x_1=\frac{2}{3}{y}_1 \\ and \frac{2y_1}{1}=\frac{3{x^2}_1+2{y^2}_1}{2} \\ \Rightarrow y_1=\frac{6}{5} \\ \Rightarrow x_1=\frac{4}{5} \\ \therefore Mid point=(\frac{4}{5},\frac{6}{5}) \\ Or \\ use the formula \\ mid point of chord lx+my+n=0 w.r.to ellipse \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 is (\frac{-a^2\ln }{a^2{l}^2+b^2{m}^2},\frac{-b^{2}mn}{a^2{l}^2+b^2{m}^2}) \end{gathered}$$