The energy of one mole of photons of radiation of frequency $$2 \times 10^{12} \mathrm{~Hz}$$ in $$\mathrm{J} ~\mathrm{mol}^{-1}$$ is ___________. (Nearest integer)

[Given : $$\mathrm{h}=6.626 \times 10^{-34} ~\mathrm{Js}$$

$$\mathrm{N}_{\mathrm{A}}=6.022 \times 10^{23} \mathrm{~mol}^{-1}$$]

Among the statements :

$$(\mathrm{S} 1)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow(\mathrm{p} \Rightarrow \mathrm{r})$$

$$(\mathrm{S} 2)~((\mathrm{p} \vee \mathrm{q}) \Rightarrow \mathrm{r}) \Leftrightarrow((\mathrm{p} \Rightarrow \mathrm{r}) \vee(\mathrm{q} \Rightarrow \mathrm{r}))$$

If $${a_n} = {{ - 2} \over {4{n^2} - 16n + 15}}$$, then $${a_1} + {a_2}\, + \,....\, + \,{a_{25}}$$ is equal to :

A straight line cuts off the intercepts $$\mathrm{OA}=\mathrm{a}$$ and $$\mathrm{OB}=\mathrm{b}$$ on the positive directions of $$x$$-axis and $$y$$ axis respectively. If the perpendicular from origin $$O$$ to this line makes an angle of $$\frac{\pi}{6}$$ with positive direction of $$y$$-axis and the area of $$\triangle \mathrm{OAB}$$ is $$\frac{98}{3} \sqrt{3}$$, then $$\mathrm{a}^{2}-\mathrm{b}^{2}$$ is equal to :