The general solution of $\frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{x+y+1}{x+y-1}$ is

There are three events $\mathrm{A}, \mathrm{B}, \mathrm{C}$, one of which must and only one can happen. The odds are 8:3 against $\mathrm{A}, 5: 2$ against B and the odds against C is $43: 17 \mathrm{k}$, then value of k is

Let $\bar{a}, \bar{b}$ and $\bar{c}$ be three non-zero vectors such that no two of them are collinear and $(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times \overline{\mathrm{c}}=\frac{1}{3}|\overline{\mathrm{~b}}||\overline{\mathrm{c}}| \overline{\mathrm{a}}$. If $\theta$ is the angle between vectors $\bar{b}$ and $\bar{c}$, then the value of $\operatorname{cosec} \theta$ is

If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=\pi$ and $x^2+y^2+z^2+k x y z=1$, then k is