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 respectively. Let the straight line $$x + y = 3$$ touch the curves C, $$E_1$$ and $$E_2$$ at $$P(x_1, y_1)$$, $$Q(x_2, y_2)$$ and $$R(x_3, y_3)$$ respectively. Given that P is the mid-point of the line segment QR and $$PQ = \dfrac{2\sqrt{2}}{3}$$, the value of $$9(x_1 y_1 + x_2 y_2 + x_3 y_3)$$ is equal to ________.

The given circle is $$x^{2}+(y-1)^{2}=2$$.
Its centre is $$O(0,1)$$ and its radius is $$\sqrt{2}$$.

The straight line is $$x+y=3 \;\; \Longleftrightarrow \;\; x+y-3=0$$.
Distance of the centre $$O(0,1)$$ from this line is $$\frac{|0+1-3|}{\sqrt{1^{2}+1^{2}}}= \frac{2}{\sqrt{2}}=\sqrt{2}$$, that is exactly the radius, hence the line is tangent to the circle.

For a circle, the point of contact is the foot of the perpendicular drawn from the centre to the tangent. For the line $$x+y-3=0$$ the foot of the perpendicular from $$(x_{0},y_{0})$$ is $$\left(x_{0}-a\frac{ax_{0}+by_{0}+c}{a^{2}+b^{2}},\;y_{0}-b\frac{ax_{0}+by_{0}+c}{a^{2}+b^{2}}\right)$$ with $$a=1,\;b=1,\;c=-3,\;(x_{0},y_{0})=(0,1).$$ Thus

$$x_{1}=0-\frac{1(-2)}{2}=1, \qquad y_{1}=1-\frac{1(-2)}{2}=2.$$

Therefore $$P(1,2)$$ and $$x_{1}y_{1}=2.$$

Since the same straight line touches the two ellipses $$E_{1},E_{2}$$ at $$Q(x_{2},y_{2})$$ and $$R(x_{3},y_{3})$$, both points lie on the line: $$x_{2}+y_{2}=3, \qquad x_{3}+y_{3}=3 \; -(1)$$

The midpoint condition $$P$$ is the midpoint of $$QR$$:

$$\frac{x_{2}+x_{3}}{2}=1, \qquad \frac{y_{2}+y_{3}}{2}=2 \; -(2)$$

Let $$x_{2}=u \;\; (\Rightarrow y_{2}=3-u)$$. From $$(2)$$, $$x_{3}=2-u$$ and with $$(1)$$, $$y_{3}=3-(2-u)=1+u.$$ Hence

$$Q(u,\,3-u), \qquad R(2-u,\,1+u).$$

The distance $$PQ$$ is given to be $$\dfrac{2\sqrt{2}}{3}$$. Compute $$PQ^{2}:$$

$$PQ^{2}=(u-1)^{2}+\bigl((3-u)-2\bigr)^{2} =(u-1)^{2}+(1-u)^{2}=2(u-1)^{2}.$$

Therefore $$\sqrt{2}\,|u-1|=\frac{2\sqrt{2}}{3}\;\Longrightarrow\;|u-1|=\frac{2}{3}.$$ So

$$u=1+\frac{2}{3}=\frac{5}{3}\quad\text{or}\quad u=1-\frac{2}{3}=\frac{1}{3}.$$

This merely swaps $$Q$$ and $$R$$; either choice gives the same final value. Choose $$u=\dfrac{5}{3}$$ (the other works identically):

$$Q\!\left(\frac{5}{3},\,\frac{4}{3}\right), \qquad R\!\left(\frac{1}{3},\,\frac{8}{3}\right).$$

Now compute the required sum:

$$x_{1}y_{1}=1\cdot2=2,$$ $$x_{2}y_{2}=\frac{5}{3}\cdot\frac{4}{3}=\frac{20}{9},$$ $$x_{3}y_{3}=\frac{1}{3}\cdot\frac{8}{3}=\frac{8}{9}.$$

Hence $$x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}=2+\frac{20}{9}+\frac{8}{9}=2+\frac{28}{9}=\frac{46}{9}.$$

Finally, $$9\bigl(x_{1}y_{1}+x_{2}y_{2}+x_{3}y_{3}\bigr)=9\cdot\frac{46}{9}=46.$$

The desired value is $$46$$.