The set of all $$\alpha$$, for which the vectors $$\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$$ and $$\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$$ are inclined at an obtuse angle for all $$t \in \mathbb{R}$$, is

Let the circles $$C_1:(x-\alpha)^2+(y-\beta)^2=r_1^2$$ and $$C_2:(x-8)^2+\left(y-\frac{15}{2}\right)^2=r_2^2$$ touch each other externally at the point $$(6,6)$$. If the point $$(6,6)$$ divides the line segment joining the centres of the circles $$C_1$$ and $$C_2$$ internally in the ratio $$2: 1$$, then $$(\alpha+\beta)+4\left(r_1^2+r_2^2\right)$$ equals

Let $$f(x)$$ be a positive function such that the area bounded by $$y=f(x), y=0$$ from $$x=0$$ to $$x=a>0$$ is $$e^{-a}+4 a^2+a-1$$. Then the differential equation, whose general solution is $$y=c_1 f(x)+c_2$$, where $$c_1$$ and $$c_2$$ are arbitrary constants, is

Let $$H: \frac{-x^2}{a^2}+\frac{y^2}{b^2}=1$$ be the hyperbola, whose eccentricity is $$\sqrt{3}$$ and the length of the latus rectum is $$4 \sqrt{3}$$. Suppose the point $$(\alpha, 6), \alpha>0$$ lies on $$H$$. If $$\beta$$ is the product of the focal distances of the point $$(\alpha, 6)$$, then $$\alpha^2+\beta$$ is equal to