Let the solution curve of the differential equation $$x \mathrm{~d} y=\left(\sqrt{x^{2}+y^{2}}+y\right) \mathrm{d} x, x>0$$, intersect the line $$x=1$$ at $$y=0$$ and the line $$x=2$$ at $$y=\alpha$$. Then the value of $$\alpha$$ is :

Considering only the principal values of the inverse trigonometric functions, the domain of the function $$f(x)=\cos ^{-1}\left(\frac{x^{2}-4 x+2}{x^{2}+3}\right)$$ is :

Let the vectors $$\vec{a}=(1+t) \hat{i}+(1-t) \hat{j}+\hat{k}, \vec{b}=(1-t) \hat{i}+(1+t) \hat{j}+2 \hat{k}$$ and $$\vec{c}=t \hat{i}-t \hat{j}+\hat{k}, t \in \mathbf{R}$$ be such that for $$\alpha, \beta, \gamma \in \mathbf{R}, \alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=\overrightarrow{0} \Rightarrow \alpha=\beta=\gamma=0$$. Then, the set of all values of $$t$$ is :

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $$\cos ^{-1}(x)-2 \sin ^{-1}(x)=\cos ^{-1}(2 x)$$ is equal to :