The shortest distance between the lines $$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$$ and $$\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$$ is

If $$\bar{a}, \bar{b}$$ and $$\bar{c}$$ are any three non-zero vectors, then $$(\bar{a}+2 \bar{b}+\bar{c}) \cdot[(\bar{a}-\bar{b}) \times(\bar{a}-\bar{b}-\bar{c})]=$$

In $$\triangle \mathrm{ABC}$$, with usual notations, $$\mathrm{m} \angle \mathrm{C}=\frac{\pi}{2}$$, if $$\tan \left(\frac{A}{2}\right)$$ and $$\tan \left(\frac{B}{2}\right)$$ are the roots of the equation $$a_1 x^2+b_1 x+c_1=0\left(a_1 \neq 0\right)$$, then

If $$\int \frac{\sin x}{3+4 \cos ^2 x} \mathrm{~d} x=\mathrm{A} \tan ^{-1}(\mathrm{~B} \cos x)+\mathrm{c}$$, (where $$\mathrm{c}$$ is a constant of integration), then the value of $$\mathrm{A}+\mathrm{B}$$ is