For $$\mathrm{t} \in(0,2 \pi)$$, if $$\mathrm{ABC}$$ is an equilateral triangle with vertices $$\mathrm{A}(\sin t,-\cos \mathrm{t}), \mathrm{B}(\operatorname{cost}, \sin t)$$ and $$C(a, b)$$ such that its orthocentre lies on a circle with centre $$\left(1, \frac{1}{3}\right)$$, then $$\left(a^{2}-b^{2}\right)$$ is equal to :

For $$\alpha \in \mathbf{N}$$, consider a relation $$\mathrm{R}$$ on $$\mathbf{N}$$ given by $$\mathrm{R}=\{(x, y): 3 x+\alpha y$$ is a multiple of 7$$\}$$. The relation $$R$$ is an equivalence relation if and only if :

Out of $$60 \%$$ female and $$40 \%$$ male candidates appearing in an exam, $$60 \%$$ candidates qualify it. The number of females qualifying the exam is twice the number of males qualifying it. A candidate is randomly chosen from the qualified candidates. The probability, that the chosen candidate is a female, is :

If $$y=y(x), x \in(0, \pi / 2)$$ be the solution curve of the differential equation

$$\left(\sin ^{2} 2 x\right) \frac{d y}{d x}+\left(8 \sin ^{2} 2 x+2 \sin 4 x\right) y=2 \mathrm{e}^{-4 x}(2 \sin 2 x+\cos 2 x)$$,

with $$y(\pi / 4)=\mathrm{e}^{-\pi}$$, then $$y(\pi / 6)$$ is equal to :