QJEEM25Ma8f09: JEE Main 2025 Mathematics - Class 11 JEEM25Ma8f09 Solution: Let and , be the vertices of a triangle. If and be its…

All the three points $A, B, C$ lie on the circle $x^2+y^2=100$ so circumcentre is $(0,0)$

$\begin{aligned}& \frac{a+0}{3}=h \Rightarrow a=3 h \\& \text { and } \frac{9+0}{3}=k \Rightarrow k=3 \\& \text { also centroid } \frac{6+10 \cos \alpha-10 \sin \alpha}{3}=h \\& \Rightarrow 10(\cos \alpha-\sin \alpha)=3 h-6\quad\text{.... (i)}\end{aligned}$

$\begin{aligned}&\begin{aligned}& \text { and } \frac{8+10 \cos \alpha-10 \sin \alpha}{3}=\mathrm{k} \\& \Rightarrow 10(\cos \alpha-\sin \alpha)=3 \mathrm{k}-8=9-8=1 \quad\text{.... (ii)}\\& \text { on squaring } 100(1-\sin 2 \alpha)=1 \\& \Rightarrow 100 \sin 2 \alpha=99\end{aligned}\\&\text { from equ. (i) and (ii) we get } \mathrm{h}=\frac{7}{3}\\&\begin{aligned}& \text { Now } 5 a-3 h+6 k+100 \sin 2 \alpha \\& =15 h-3 h+6 k+100 \sin 2 \alpha \\& =12 \times \frac{7}{3}+18+99 \\& =145\end{aligned}\end{aligned}$