Match the column. Normals at P, Q, R are drawn to y^2=4 x which...

Since, equation of normal is $y+x t=2 a t+a t^3$ passes through $(3,0)$. \Rightarrow \quad 3 t=2 t+t³ \Rightarrow t=0,1,-1 $\text{. }\therefore P(1,2), Q(0,0), R(1,-2)$. Thus, (i) Area of $\triangle P Q R=\frac{1}{2} \times 1 \times 4=2$ (ii) Centroid of \triangle P Q R= \left(\frac{2}{3},0\right) Equation of circle passing through $P, Q, R$ is $$\begin{aligned} & \quad(x-1)(x-1)+(y-2)(y+2)+\lambda(x-1)=0 \\ & \Rightarrow \quad 1-4-\lambda=0 \Rightarrow \lambda=-3 \\ & \therefore \text{ Required equation of circle is } x^2+y²-5 x-1=0 \\ & \therefore \text{ Centre } \ left(\frac{5}{2},0)\text{ and radius } \frac{5}{2}. \end{aligned}$$$$