Let the values of $\lambda$ for which the shortest distance between the lines $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$
and $\frac{x-\lambda}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_1$ and $\lambda_2$. Then the radius of the circle passing through the
points $(0, 0), (\lambda_1, \lambda_2)$ and $(\lambda_2, \lambda_1)$ is

Let $f(x) = x - 1$ and $g(x) = e^x$ for $x \in \mathbb{R}$. If $\frac{dy}{dx} = \left( e^{-2\sqrt{x}} g\left(f(f(x))\right) - \frac{y}{\sqrt{x}} \right)$, $y(0) = 0$, then $y(1)$ is

If $ \frac{1}{1^4} + \frac{1}{2^4} + \frac{1}{3^4} + \ldots \infty= \frac{\pi^4}{90} $,

$\frac{1}{1^4} + \frac{1}{3^4} + \frac{1}{5^4} + \ldots \infty= \alpha $,

$ \frac{1}{2^4} + \frac{1}{4^4} + \frac{1}{6^4} + \ldots \infty= \beta $,

then $ \frac{\alpha}{\beta} $ is equal to :

Let a be the length of a side of a square OABC with O being the origin. Its side OA makes an acute angle $$\alpha $$ with the positive x-axis and the equations of its diagonals are $(\sqrt{3}+1)x+(\sqrt{3}-1)y=0$ and $(\sqrt{3}-1)x-(\sqrt{3}+1)y+8\sqrt{3}=0$. Then $a$² is equal to :