All the three points $A, B, C$ lie on the circle $x^2+y^2=100$ so circumcentre is $(0,0)$ $\begin{aligned} & \frac{a+0}{3}=h \Rightarrow a=3 h \\ & \text { and } \frac{9+0}{3}=k \Rightarrow k=3 \\ & \text { also centroid } \frac{6+10 \cos \alpha-10 \sin \alpha}{3}=h \\ & \Rightarrow 10(\cos \alpha-\sin \alpha)=3 h-6\quad\text{.... (i)} \end{aligned}$ $\begin{aligned} &\begin{aligned} & \text { and } \frac{8+10 \cos \alpha-10 \sin \alpha}{3}=\mathrm{k} \\ & \Rightarrow 10(\cos \alpha-\sin \alpha)=3 \mathrm{k}-8=9-8=1 \quad\text{.... (ii)}\\ & \text { on squaring } 100(1-\sin 2 \alpha)=1 \\ & \Rightarrow 100 \sin 2 \alpha=99 \end{aligned}\\ &\text { from equ. (i) and (ii) we get } \mathrm{h}=\frac{7}{3}\\ &\begin{aligned} & \text { Now } 5 a-3 h+6 k+100 \sin 2 \alpha \\ & =15 h-3 h+6 k+100 \sin 2 \alpha \\ & =12 \times \frac{7}{3}+18+99 \\ & =145 \end{aligned} \end{aligned}$