$$ \cot ^{-1}\left(2 \cdot 1^2\right)+\cot ^{-1}\left(2 \cdot 2^2\right)+\cot ^{-1}\left(2 \cdot 3^2\right)+\ldots \ldots \ldots \infty= $$

With usual notation, in a triangle ABC $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$, then the value of $\cos B$ is equal to

If $\overline{\mathrm{a}}=\frac{1}{\sqrt{10}}(3 \hat{\mathrm{i}}+\hat{\mathrm{k}}), \overline{\mathrm{b}}=\frac{1}{7}(2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}})$, then the value of $(\overline{\mathrm{a}}-2 \overline{\mathrm{~b}}) \cdot\{(\overline{\mathrm{a}} \times \overline{\mathrm{b}}) \times(2 \overline{\mathrm{a}}+\overline{\mathrm{b}})\}$ is

The vectors $\bar{p}=\hat{i}+a \hat{j}+a^2 \hat{k}, \bar{q}=\hat{i}+b \hat{j}+b^2 \hat{k}$ and $\overline{\mathrm{r}}=\hat{\mathrm{i}}+\mathrm{c} \hat{\mathrm{j}}+\mathrm{c}^2 \hat{\mathrm{k}}$ are non-coplanar and $\left|\begin{array}{lll}a & a^2 & 1+a^3 \\ b & b^2 & 1+b^3 \\ c & c^2 & 1+c^3\end{array}\right|=0$ then the value of $(a b c)$ is