If G be the centroid and I be the incentre of the triangle with vertices $A\left(-36,7\right),B\left(20,7\right)$ and $C\left(0,-8\right)$ and $GI=\frac{25}{3}\sqrt{205}\lambda$ the $\lambda$ is

$\begin{array}{l}G=\left(\frac{-36+20+0}{3},\frac{7+7-8}{3}\right)=\left(\frac{-16}{3},2\right)a=BC=\sqrt{(0-20{)}^2+(-8-7{)}^2}=25\\ b=CA=\sqrt{{36}^2+(-8-7{)}^2}=39,c=AB=\sqrt{(-36-20{)}^2+(7-7{)}^2}=56\\ \text{ Let }I=(\alpha ,\beta ),\text{ then }\alpha =\frac{-36\times 25+39\times 20+56\times 0}{25+39+56}=-1\beta =\frac{25\times 7+39\times 7-8\times 56}{25+39+56}=0.\\ I=(-1,0)\cdot GI=\sqrt{{\left(\frac{-16}{3}+1\right)}^2+(2-0{)}^2}=\frac{1}{3}\sqrt{205}=\frac{25}{3}\sqrt{205}\lambda \Rightarrow \lambda =\frac{1}{25}.\end{array}$