The shortest distance between the lines $$\frac{x-3}{4}=\frac{y+7}{-11}=\frac{z-1}{5}$$ and $$\frac{x-5}{3}=\frac{y-9}{-6}=\frac{z+2}{1}$$ is:

A variable line $$\mathrm{L}$$ passes through the point $$(3,5)$$ and intersects the positive coordinate axes at the points $$\mathrm{A}$$ and $$\mathrm{B}$$. The minimum area of the triangle $$\mathrm{OAB}$$, where $$\mathrm{O}$$ is the origin, is :

The solution curve, of the differential equation $$2 y \frac{\mathrm{d} y}{\mathrm{~d} x}+3=5 \frac{\mathrm{d} y}{\mathrm{~d} x}$$, passing through the point $$(0,1)$$ is a conic, whose vertex lies on the line :

Let $$\lambda, \mu \in \mathbf{R}$$. If the system of equations

$$\begin{aligned} & 3 x+5 y+\lambda z=3 \\ & 7 x+11 y-9 z=2 \\ & 97 x+155 y-189 z=\mu \end{aligned}$$

has infinitely many solutions, then $$\mu+2 \lambda$$ is equal to :