Let $E_1$ and $E_2$ be two ellipses whose centers are at the origin. The major axes of   $E_1$ and $E_2$  lie along the x-axis and the y-axis, respectively. Let S be the circle $x^2+{\left(y-1\right)}^2=2.$ The straight line $x+y=3$ touches the curves S,  $E_1$ and $E_2$ at P, Q and R, respectively. Suppose that $PQ=PR=\frac{2\sqrt{2}}{3}.$ If  $e_1$and  $e_2$are the eccentricities of   $E_1$ and $E_2$   respectively, then the correct expression(s) is (are)

The line $x+y=3$ touches the circle $x^2+{\left(y-1\right)}^2=2$ at P

P is foot of perpendicular from $\left(0,1\right)$ to $x+y=3$

$\Rightarrow$ $P\left(1,2\right)$

Since $PQ=\frac{2\sqrt{2}}{3}=PR$  we have $Q\left(1+\frac{2\sqrt{2}}{3\sqrt{2}},2-\frac{2\sqrt{2}}{3\sqrt{2}}\right)$

$E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $e_1^2=\frac{a^2-b^2}{a^2}$

$\therefore a^2+b^2=9\left(x+y=3touches{E}_1\Rightarrow a^2{m}^2+b^2=c^2\right)$

$Since Q lies on E_1, we have \frac{25}{9a^2}+\frac{16}{9b^2}=1$

$\Rightarrow$ $a^2=5,b^2=4$

$\therefore e_1^2=\frac{1}{5}$

similarly 

$R\left(1-\frac{2}{3},2+\frac{2}{3}\right)\text{\hspace{0.17em}lies on }\frac{x^2}{c^2}+\frac{y^2}{d^2}=1c<d$

$\therefore \left.\begin{array}{c}{c}^2+d^2=9\\ \frac{1}{9c^2}+\frac{64}{9d^2}=1\end{array}\right\}solving{c}^2=1,d^2=8$

hence, $\therefore e_2^2=\frac{7}{8}$

Verifying the given options, we can see that options (A),(B) are correct.