Let A(6,8),$$B(10\cos \alpha, -10\sin \alpha)$$ and $$C(-10\sin \alpha, 10\cos \alpha)$$. be the vertices of a triangle. If L(a, 9) and G(h, k) be its orthocenter and centroid respectively, then $$(5a − 3h + 6k + 100 \sin 2\alpha)$$ is equals to_______.

Given $$A(6,8)$$, $$B(10\cos\alpha,-10\sin\alpha)$$, $$C(-10\sin\alpha,10\cos\alpha)$$, orthocentre $$L(a,9)$$ and centroid $$G(h,k)$$, we seek to find $$5a-3h+6k+100\sin2\alpha$$.

Note that $$B$$ and $$C$$ lie on a circle of radius 10 centred at the origin since $$|OB|^2 = 100\cos^2\alpha + 100\sin^2\alpha = 100$$ and $$|OA|^2 = 36 + 64 = 100$$.

For a triangle inscribed in a circle, the orthocentre $$L$$ and centroid $$G$$ relate to the circumcentre $$O(0,0)$$ by $$\vec{OL} = \vec{OA} + \vec{OB} + \vec{OC}$$. Thus $$L = A + B + C = (6+10\cos\alpha-10\sin\alpha,\;8-10\sin\alpha+10\cos\alpha),$$ so $$a = 6+10\cos\alpha-10\sin\alpha$$. From $$L=(a,9)$$ we have $$8-10\sin\alpha+10\cos\alpha = 9$$ and hence $$10\cos\alpha-10\sin\alpha = 1$$.

It follows that $$a = 6 + (10\cos\alpha-10\sin\alpha) = 6 + 1 = 7.$$

The centroid is $$G = \frac{A+B+C}{3} = \frac{L}{3} = \Bigl(\frac{6+10\cos\alpha-10\sin\alpha}{3},\frac{8-10\sin\alpha+10\cos\alpha}{3}\Bigr) = \Bigl(\frac{7}{3},3\Bigr),$$ so $$h = \tfrac{7}{3},\quad k = 3.$$

Squaring the relation $$10\cos\alpha-10\sin\alpha = 1$$ gives $$100 - 200\sin\alpha\cos\alpha = 1$$ and therefore $$100\sin2\alpha = 99.$$

Finally, $$5a-3h+6k+100\sin2\alpha = 5(7) - 3\Bigl(\frac{7}{3}\Bigr) + 6(3) + 99 = 35 - 7 + 18 + 99 = 145.$$

The correct answer is 145.