If the angles of a $$\triangle A B C$$ are in the ratio $$1: 2: 3$$, then the corresponding sides are in the ratio

In a $$\triangle A B C, r_1 \cot \frac{A}{2}+r_2 \cot \frac{B}{2}+r_3 \cot \frac{C}{2}=$$

The point of intersection of the lines $$\mathbf{r}=2 \mathbf{b}+t(6 \mathbf{c}-\mathbf{a})$$ and $$\mathbf{r}=\mathbf{a}+s(\mathbf{b}-3 \mathbf{c})$$ is

In quadrilateral $$A B C D, \mathbf{A B}=\mathbf{a}, \mathbf{B C}=\mathbf{b}$$. $$\mathbf{D A}=\mathbf{a}-\mathbf{b}, M$$ is the mid-point of $$B C$$ and $$X$$ is a point on DM such that, $$\mathbf{D X}=\frac{4}{5}$$ DM. Then, the points $$A, X$$ and $$C$$.