Let C be the circle x^2+( y-1)^2=2, E_1 and E_2 be two ellipses whose centres lie at the origin and major axes lie on x -axis and y -axis…

Let $\mathrm{E}_1: \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1,(\mathrm{a} \gt \mathrm{b})$E1​:a2x2​+ b2y2​=1,(a>b)

$$\mathrm{E}_2: \frac{\mathrm{x}^2}{\mathrm{c}^2}+\frac{\mathrm{y}^2}{\mathrm{~d}^2}=1,(\mathrm{c} \lt \mathrm{d})$$E2​:c2x2​+ d2y2​=1,(c<d)

$$C: x^2+(y-1)^2=2$$C:x2+(y−1)2=2

Equation of tangent at $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$P(x1​,y1​)

$$\mathrm{xx}_1+\mathrm{y}\left(\mathrm{y}_1-1\right)=\left(\mathrm{y}_1+1\right)$$xx1​+y(y1​−1)=(y1​+1)

comparing with $\mathrm{x}+\mathrm{y}=3$x+y=3 we get $\mathrm{P}(1,2)$P(1,2)

$\because$∵ Now parametric equation of $x+y=3$x+y=3

$$\frac{(\mathrm{x}-1)}{\left(\frac{-1}{\sqrt{2}}\right)}=\frac{(\mathrm{y}-2)}{\left(\frac{1}{\sqrt{2}}\right)}= \pm \frac{2 \sqrt{2}}{3} \quad\left(\because \mathrm{PQ}=\frac{2 \sqrt{2}}{3}\right)$$(2​−1​)(x−1)​=(2​1​)(y−2)​=±322​​(∵PQ=322​​)

On solving we get $\mathrm{Q}\left(\frac{5}{3}, \frac{4}{3}\right), \mathrm{R}\left(\frac{1}{3}, \frac{8}{3}\right)$Q(35​,34​),R(31​,38​)

So, $9\left(\mathrm{x}_1 \mathrm{y}_1+\mathrm{x}_2 \mathrm{y}_2+\mathrm{x}_3 \mathrm{y}_3\right)$9(x1​y1​+x2​y2​+x3​y3​)

$\begin{aligned}
& 9\left(2+\frac{5}{3} \times \frac{4}{3}+\frac{1}{3} \times \frac{8}{3}\right) \
& \Rightarrow 46
\end{aligned}$