The general solution of the differential equation $$\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$$ is :

A line, with the slope greater than one, passes through the point $$A(4,3)$$ and intersects the line $$x-y-2=0$$ at the point B. If the length of the line segment $$A B$$ is $$\frac{\sqrt{29}}{3}$$, then $$B$$ also lies on the line :

Let the locus of the centre $$(\alpha, \beta), \beta>0$$, of the circle which touches the circle $$x^{2}+(y-1)^{2}=1$$ externally and also touches the $$x$$-axis be $$\mathrm{L}$$. Then the area bounded by $$\mathrm{L}$$ and the line $$y=4$$ is:

Let $$\mathrm{P}$$ be the plane containing the straight line $$\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$$ and perpendicular to the plane containing the straight lines $$\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$$ and $$\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$$. If $$\mathrm{d}$$ is the distance of $$\mathrm{P}$$ from the point $$(2,-5,11)$$, then $$\mathrm{d}^{2}$$ is equal to :