If the solution curve, of the differential equation $$\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y-2}{x-y}$$ passing through the point $$(2,1)$$ is $$\tan ^{-1}\left(\frac{y-1}{x-1}\right)-\frac{1}{\beta} \log _{\mathrm{e}}\left(\alpha+\left(\frac{y-1}{x-1}\right)^2\right)=\log _{\mathrm{e}}|x-1|$$, then $$5 \beta+\alpha$$ is equal to __________.

If $$y=y(x)$$ is the solution of the differential equation

$$\frac{d y}{d x}+\frac{4 x}{\left(x^{2}-1\right)} y=\frac{x+2}{\left(x^{2}-1\right)^{\frac{5}{2}}}, x > 1$$ such that

$$y(2)=\frac{2}{9} \log _{e}(2+\sqrt{3}) \text { and } y(\sqrt{2})=\alpha \log _{e}(\sqrt{\alpha}+\beta)+\beta-\sqrt{\gamma}, \alpha, \beta, \gamma \in \mathbb{N} \text {, then } \alpha \beta \gamma \text { is equal to }$$ :

Let the tangent at any point P on a curve passing through the points (1, 1) and $$\left(\frac{1}{10}, 100\right)$$, intersect positive $$x$$-axis and $$y$$-axis at the points A and B respectively. If $$\mathrm{PA}: \mathrm{PB}=1: k$$ and $$y=y(x)$$ is the solution of the differential equation $$e^{\frac{d y}{d x}}=k x+\frac{k}{2}, y(0)=k$$, then $$4 y(1)-6 \log _{\mathrm{e}} 3$$ is equal to ____________.