$$\text { If } \log (x+y)=2 x y \text {, then } \frac{\mathrm{d} y}{\mathrm{~d} x} \text { at } x=0 \text { is }$$

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