Let $R$ be the focus of the parabola $y^{2}=20 x$ and the line $y=m x+c$ intersect the parabola at two points $P$ and $Q$.

Let the point $G(10,10)$ be the centroid of the triangle $P Q R$. If $c-m=6$, then $(P Q)^{2}$ is :

$y^2=20 x, y=m x+\mathrm{c}$ <br/><br/>Put value of $x$ <br/><br/>$$ \begin{aligned} & y^2=20\left(\frac{y-c}{m}\right) \\\\ & \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 .......(i) \end{aligned} $$ <br/><br/>Since, centroid $=(10,10)$ <br/><br/>$$ \begin{aligned} & \text { So, } \frac{y_1+y_2+0}{3}=10 \\\\ & \Rightarrow y_1+y_2=30 \end{aligned} $$ <br/><br/>From (1), <br/><br/>$\text { Sum of roots }=\frac{20}{m}=30 \Rightarrow m=\frac{2}{3}$ <br/><br/>Also, $c-m=6 \Rightarrow c=6+\frac{2}{3}=\frac{20}{3}$ <br/><br/>Now, the equation is : <br/><br/>$$ \begin{aligned} & y^2-\frac{20}{2} \times 3 y+\frac{20}{2} \times 3 \times \frac{20}{3}=0 \\\\ & \Rightarrow y^2-30 y+200=0 \\\\ & \Rightarrow y^2-20 y-10 y+200=0 \\\\ & \Rightarrow(y-20)(y-10)=0 \\\\ & \Rightarrow y=10,20 \Rightarrow x=5, x=20 \\\\ & \therefore P \equiv(5,10), Q \equiv(20,20) \\\\ & \text { So, }(P Q)^2=(20-5)^2+(20-10)^2 \\\\ & =225+100=325 \end{aligned} $$