Let $A, B, C$ be three points in xy-plane, whose position vector are given by $\sqrt{3} \hat{i}+\hat{j}, \hat{i}+\sqrt{3} \hat{j}$ and $a \hat{i}+(1-a) \hat{j}$ respectively with respect to the origin O . If the distance of the point C from the line bisecting the angle between the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ is $\frac{9}{\sqrt{2}}$, then the sum of all the possible values of $a$ is :

If $\sum\limits_{r=1}^{13}\left\{\frac{1}{\sin \left(\frac{\pi}{4}+(r-1) \frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}+\frac{r \pi}{6}\right)}\right\}=a \sqrt{3}+b, a, b \in Z$, then $a^2+b^2$ is equal to :

If $f(x)=\int \frac{1}{x^{1 / 4}\left(1+x^{1 / 4}\right)} \mathrm{d} x, f(0)=-6$, then $f(1)$ is equal to :