Given, circle is $$x^2+y^2-2 x-4 y=0$$ and parabola $$y^2=8 x$$.

$$\because$$ Both the curves intersect each other at P.

$$\because \quad x^2+8 x-2 x-4 \cdot 2 \sqrt{2 x}=0$$

$$\begin{array}{lr} \Rightarrow & x^2+6 x-8 \sqrt{2 x}=0 \\ \Rightarrow & \sqrt{x}\left[x^{\frac{3}{2}}+6 x^{\frac{1}{2}}-8 \sqrt{2}\right]=0 \\ \Rightarrow & \text { Let } \sqrt{x}=t \\ \therefore & t\left[t^3+6 t-8 \sqrt{2}\right]=0 \\ \Rightarrow & t(t-\sqrt{2})\left(t^2-\sqrt{2} t+4\right)=0 \\ \Rightarrow & t=0 \text { or } t=\sqrt{2} \text { or } t=\frac{\sqrt{2} \pm \sqrt{2-4(4)}}{2} \end{array}$$

(rejected because it is imaginary)

$$\begin{array}{ll} \Rightarrow t=0 & \text { or } t=\sqrt{2} \\ \Rightarrow x=0 & \text { or } x=2 \\ \Rightarrow y=0 & \text { or } y=4 \end{array}$$

Hence, the required coordinates are $$\mathrm{P}(2,4), Q(0,0)$$ and $$S(2,0)$$.

$$\therefore \quad$$ Area of $$\triangle \mathrm{PQS}=\frac{1}{2} \times 2 \times 4=4$$