Let the foot of perpendicular from the point $(\lambda, 2,3)$ on the line $\frac{x-4}{1}=\frac{y-9}{2}=\frac{z-5}{1}$ be the point ( $1, \mu, 2$ ). Then the distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6}$ and $\frac{x-\lambda}{2}=\frac{y-\mu}{3}=\frac{z+5}{6}$ is equal to :

The value of the integral $\int\limits_0^2 \frac{\sqrt{x\left(x^2+x+1\right)}}{(\sqrt{x+1})\left(\sqrt{x^4+x^2+1}\right)} \mathrm{d} x$ is equal to:

Let $y=y(x)$ be the solution of the differential equation $x \sqrt{1-x^2} d y+\left(y \sqrt{1-x^2}-x \cos ^{-1} x\right) d x=0, x \in(0,1), \lim _{x \rightarrow 1^{-}} y(x)=1$. Then $y\left(\frac{1}{2}\right)$ equals :

Let $f:(1, \infty) \rightarrow \mathbf{R}$ be a function defined as $f(x)=\frac{x-1}{x+1}$. Let $f^{i+1}(x)=f\left(f^i(x)\right), i=1,2, \ldots, 25$, where $f^1(x)=f(x)$. If $g(x)+f^{26}(x)=0, x \in(1, \infty)$, then the area of the region bounded by the curves $y=g(x), 2 y=2 x-3, y=0$ and $x=4$ is :