Let $$\mathrm{A}$$ be the point of intersection of the lines $$3 x+2 y=14,5 x-y=6$$ and $$\mathrm{B}$$ be the point of intersection of the lines $$4 x+3 y=8,6 x+y=5$$. The distance of the point $$P(5,-2)$$ from the line $$\mathrm{AB}$$ is

If $$\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C$$, where $$C$$ is the integration constant, then $$A B$$ is equal to

The distance of the point $$(2,3)$$ from the line $$2 x-3 y+28=0$$, measured parallel to the line $$\sqrt{3} x-y+1=0$$, is equal to

Let a unit vector $$\hat{u}=x \hat{i}+y \hat{j}+z \hat{k}$$ make angles $$\frac{\pi}{2}, \frac{\pi}{3}$$ and $$\frac{2 \pi}{3}$$ with the vectors $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}, \frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ and $$\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}$$ respectively. If $$\vec{v}=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{j}+\frac{1}{\sqrt{2}} \hat{k}$$ then $$|\hat{u}-\vec{v}|^2$$ is equal to