If the shortest distance of the parabola $y^2=4 x$ from the centre of the circle $x^2+y^2-4 x-16 y+64=0$ is $\mathrm{d}$, then $\mathrm{d}^2$ is equal to :

If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x, \mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$, then $36 \frac{\mathrm{I}_1}{\mathrm{I}_2}$ is equal to :

Let $x=x(\mathrm{t})$ and $y=y(\mathrm{t})$ be solutions of the differential equations $\frac{\mathrm{d} x}{\mathrm{dt}}+\mathrm{a} x=0$ and $\frac{\mathrm{d} y}{\mathrm{dt}}+\mathrm{by}=0$ respectively, $\mathrm{a}, \mathrm{b} \in \mathbf{R}$. Given that $x(0)=2 ; y(0)=1$ and $3 y(1)=2 x(1)$, the value of $\mathrm{t}$, for which $x(\mathrm{t})=y(\mathrm{t})$, is :

The distance, of the point $(7,-2,11)$ from the line

$\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}$ along the line $\frac{x-5}{2}=\frac{y-1}{-3}=\frac{z-5}{6}$, is :