The mid point of the chord x + y = 2 of the Ellipse $3{\mathrm{x}}^2+2{\mathrm{y}}^2=6$ is 

$Given ellipse \frac{x^2}{2}+\frac{y^2}{3}=1$

Given line  x + y = 2 .....(1) 
let $\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is a mid point then equation of chord is
$3{\mathrm{xx}}_1+2{\mathrm{yy}}_1=3{\mathrm{x}}_1^2+2{\mathrm{y}}_1^2\dots \dots ..\left(2\right)$
Comparing (1) & (2)

$$\begin{gathered} \frac{3x_1}{1}=\frac{2y_1}{1}=\frac{3{x_1}^2+2{y_1}^2}{2} \\ \Rightarrow x_1=\frac{2}{3}{y}_1 and \\ 4y_1=3{x_1}^2+2{y_1}^2 \\ \Rightarrow 4y_1=3\left(\frac{4}{9}\right){y_1}^2+2{y_1}^2 \\ \Rightarrow 30{y^2}_1=36y_1 \\ \Rightarrow y_1=\frac{6}{5} \\ Now x_1=\frac{2}{3}\left(\frac{6}{5}\right)=\frac{4}{5} \\ So mid point of chord=(\frac{4}{5},\frac{6}{5}) \end{gathered}$$