1. Find the tangency point $P$ of the line with the circle

The circle is $x^2+(y-1)^2=2$ with centre $O_C=(0,1)$ and radius $r=\sqrt2$.

The line is $x+y=3.$ Its distance from $(0,1)$ is $\frac{|0+1-3|}{\sqrt{1^2+1^2}}=\frac{2}{\sqrt2}=\sqrt2,$ so it is indeed tangent.

For a tangent point, the radius is perpendicular to the tangent. Since the line has slope $-1$, the radius has slope $1$. So the radius through $(0,1)$ is $y=x+1.$ Intersect with $x+y=3$:

$y=2.$ Hence $P=(x_1,y_1)=(1,2).$

2. Use midpoint condition to get $R$ from $Q$

Given $P$ is midpoint of $QR$, $P=\left(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\right)=(1,2).$ Thus $x_2+x_3=2,\qquad y_2+y_3=4.$

Also $Q,R$ lie on the same line $x+y=3$, and $P$ lies on that line. So they are collinear.

Given $PQ=\frac{2\sqrt2}{3}.$ Since $P$ is midpoint of $QR$, we also have $PR=PQ=\frac{2\sqrt2}{3}.$

The direction vector of the line $x+y=3$ is $(1,-1)$, whose unit vector is $\frac{1}{\sqrt2}(1,-1).$ So moving distance $\frac{2\sqrt2}{3}$ along this line changes coordinates by $\frac{2\sqrt2}{3}\cdot \frac{1}{\sqrt2}(1,-1)=\left(\frac23,-\frac23\right).$ Hence the two points at that distance from $P=(1,2)$ on the line are $\left(1+\frac23,2-\frac23\right)=\left(\frac53,\frac43\right),$ $\left(1-\frac23,2+\frac23\right)=\left(\frac13,\frac83\right).$ So $Q$ and $R$ are these two points in some order.

3. Identify which point belongs to which ellipse

Ellipse $E_1$

Its centre is origin and major axis is along the $x$-axis, so its equation is of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b>0.$ For the tangent line $x+y=3$, write it as $y=-x+3.$ A line $y=mx+c$ is tangent to this ellipse iff $c^2=a^2m^2+b^2.$ Here $m=-1,c=3$, hence $9=a^2+b^2.$

For ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, tangent at $(x_0,y_0)$ is $\frac{xx_0}{a^2}+\frac{yy_0}{b^2}=1.$ Comparing with $x+y=3$, i.e. $\frac{x+y}{3}=1,$ we get $\frac{x_0}{a^2}=\frac13,\qquad \frac{y_0}{b^2}=\frac13,$ so $x_0=\frac{a^2}{3},\qquad y_0=\frac{b^2}{3}.$ Thus $x_0+y_0=\frac{a^2+b^2}{3}=3.$ So the tangency point on such an ellipse is $\left(\frac{a^2}{3},\frac{b^2}{3}\right),$ with first coordinate larger than second because $a>b$.

Among our two candidate points, only $\left(\frac53,\frac43\right)$ has $x>y$. Therefore $Q=\left(\frac53,\frac43\right).$

Ellipse $E_2$

Its centre is origin and major axis is along the $y$-axis, so its tangency point must have $y>x$. Thus $R=\left(\frac13,\frac83\right).$

4. Compute $x_1y_1+x_2y_2+x_3y_3$

Now $x_1y_1=1\cdot 2=2,$ $x_2y_2=\frac53\cdot \frac43=\frac{20}{9},$ $x_3y_3=\frac13\cdot \frac83=\frac{8}{9}.$ Therefore $x_1y_1+x_2y_2+x_3y_3=2+\frac{20}{9}+\frac{8}{9}=2+\frac{28}{9}=\frac{46}{9}.$ Hence $9(x_1y_1+x_2y_2+x_3y_3)=46.$

5. Final answer

The required integer is $\boxed{46}.$

This matches the stored correct answer.