JEE Main 2025 Jan 22 Shift 2, Mathematics Q22: Points and Straight Lines

Observe that $A$, $B$, $C$ all lie on the circle $x^2+y^2=100$ (check: $6^2+8^2=100$, $100{\cos }^2\alpha +100{\sin }^2\alpha =100$, similarly for $C$). So the circumcentre is $O=(0,0)$.

Use Euler's line relation: the centroid $G$ divides the segment from $O$ to the orthocentre $L$ in ratio $1:2$, i.e. $\overrightarrow{OG}=\frac{1}{3}\overrightarrow{OL}$, so $G=\frac{L}{3}=\left(\frac{a}{3},3\right)$.

Thus $h=\frac{a}{3}$ (i.e. $a=3h$) and $k=3$.

Compute the centroid directly:

$h=\frac{6+10\cos \alpha -10\sin \alpha }{3}\Rightarrow 10(\cos \alpha -\sin \alpha )=3h-6$ ... (i)

$k=\frac{8-10\sin \alpha +10\cos \alpha }{3}=3\Rightarrow 10(\cos \alpha -\sin \alpha )=9-8=1$ ... (ii)

From (i) and (ii): $3h-6=1\Rightarrow h=\frac{7}{3}$, and so $a=3h=7$.

Squaring (ii): $100(\cos \alpha -\sin \alpha {)}^2=1\Rightarrow 100(1-\sin 2\alpha )=1\Rightarrow 100\sin 2\alpha =99$.

Finally:

$5a-3h+6k+100\sin 2\alpha =5(7)-3\cdot \frac{7}{3}+6(3)+99=35-7+18+99=145$.