In a $\triangle A B C$ if $a+c=5 b$, then $\cot \frac{A}{2} \cot \frac{C}{2}=$

In a $\triangle A B C$, if $r_1=3, r_2=4, r_3=6$, then $b=$

Let the position vectors of the vertices of a $\triangle A B C$ be $\mathbf{a , b}, \mathbf{c}$. If on the plane of the triangle, $P$ is a point having position vector $\mathbf{x}$ such that $\mathbf{x} \cdot(\mathbf{c}-\mathbf{b})=\mathbf{a} \cdot \mathbf{c}-\mathbf{a} \cdot \mathbf{b}$ and $\mathbf{x} \cdot(\mathbf{a}-\mathbf{c})=\mathbf{a b}-\mathbf{b} \mathbf{c}$, then for the $\triangle A B C, P$ is the

The point of intersection of the lines represented by $\mathbf{r}=(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mathbf{t}(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$ and $\mathbf{r}=(4 \hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathbf{s}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$ is