Solving the line $x+y=3$, and the circle $x^2+$ $(y-1)^2=2$

Substitute $y=3-x$ :

$$\begin{aligned} & x^2+(3-x-1)^2=2 \\ & \Rightarrow x^2-2 x+1=0 \\ & \Rightarrow x=1 \Rightarrow y=2 \end{aligned}$$

So, $P=\left(x_1, y_1\right)=(1,2) \Rightarrow x_1 y_1=1 \cdot 2=2$

Use midpoint condition

Let $Q=\left(x_2, y_2\right), R=\left(x_3, y_3\right)$.

Since $P$ is the midpoint of QR:

$$x_2+x_3=2 x_1=2, y_2+y_3=2 y_1=4

$$So, we can write: $x_3=2-x_2, y_3=4-y_2$

Given,

$$P Q=\frac{2 \sqrt{2}}{3} \Rightarrow P Q^2=\left(x_2-1\right)^2+\left(y_2-2\right)^2=\frac{8}{9}$$

Let's denote: $x_2=a, y_2=b, x_3=2-a, y_3=4-b$

$$\begin{aligned} & (a-1)^2+(b-2)^2=\frac{8}{9} \\ & \Rightarrow a^2-2 a+1+b^2-4 b+4=\frac{8}{9} \\ & \Rightarrow a^2+b^2-2 a-4 b+5=\frac{8}{9} \\ & \Rightarrow 9 a^2+9 b^2-18 a-36 b+37=0 \end{aligned}$$

Hence, $a=\frac{5}{3}, b=\frac{4}{3}$

$$\begin{aligned} & x_1 y_1+x_2 y_2+x_3 y_3=2+a b+(2-a)(4-b) \\ & 9\left(x_1 y_1+x_2 y_2+x_3 y_3\right)=9(10+2 a b-2 b-4 a) \\ & =90+18 a b-18 b-36 a=46 \end{aligned}$$