If general solution of $$\cos ^2 \theta-2 \sin \theta+\frac{1}{4}=0$$ is $$\theta=\frac{\mathrm{n} \pi}{\mathrm{A}}+(-1)^{\mathrm{n}} \frac{\pi}{\mathrm{B}}, \mathrm{n} \in \mathrm{Z}$$, then $$\mathrm{A}+\mathrm{B}$$ has the

$$\overline{\mathrm{u}}, \overline{\mathrm{v}}, \overline{\mathrm{w}}$$ are three vectors such that $$|\overline{\mathrm{u}}|=1, |\bar{v}|=2,|\bar{w}|=3$$. If the projection of $$\bar{v}$$ along $$\bar{u}$$ is equal to projection of $$\bar{w}$$ along $$\bar{u}$$ and $$\bar{v}, \bar{w}$$ are perpendicular to each other, then $$|\bar{u}-\bar{v}+\bar{w}|=$$

The distance of the point $$\mathrm{P}(-2,4,-5)$$ from the line $$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$ is

$$\mathrm{A}$$ and $$\mathrm{B}$$ are independent events with $$\mathrm{P}(\mathrm{A})=\frac{1}{4}$$ and $$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A})$$, then $$\mathrm{P}(\mathrm{B})$$ is