Let A(6, 8), B(10 cosα, –10 sinα) and C (–10 sinα, 10 cosα), be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then (5a – 3h + 6k + 100 sin2α) is equal to___________

All the three points A, B, C lie on the circle x² + y² = 100 so circumcentre is (0, 0) 

$\frac{\mathrm{a}+0}{3}=\mathrm{h}\Rightarrow \mathrm{a}=3\mathrm{h}$ and $\frac{9+0}{3}=\mathrm{k}\Rightarrow \mathrm{k}=3$
also centroid $\frac{6+10\cos \alpha -10\sin \alpha }{3}=\mathrm{h}$
$\Rightarrow 10(\cos \mathrm{\alpha }-\sin \mathrm{\alpha })=3\mathrm{h}-6 ...\left(\mathrm{i}\right)$ and $\frac{8+10\cos \mathrm{\alpha }-10\sin \mathrm{\alpha }}{3}=\mathrm{k}$
$\Rightarrow 10(\cos \mathrm{\alpha }-\sin \mathrm{\alpha })=3\mathrm{k}-8=9-8=1 ...\left(\mathrm{ii}\right)$ 
on squaring 100(1 – sin2α) = 1
$\Rightarrow 100\sin 2\mathrm{\alpha }=99$
from equ. (i) and (ii) we get $\mathrm{h}=\frac{7}{3}$
Now 5a – 3h + 6k + 100 sin2α 
= 15h – 3h + 6k + 100 sin2α
$\begin{array}{l}=12\times \frac{7}{3}+18+99\\ =145\end{array}$