'I' is the incentre of triangle ABC whose corresponding sides are a, b, c, respectively then $\overrightarrow{\mathrm{IA}}+\mathrm{b}\overrightarrow{\mathrm{IB}}+\mathrm{c}\overrightarrow{\mathrm{IC}}$ is always equal to

Let the incentre be at the origin and be $\mathrm{A}\left(\overrightarrow{\mathrm{p}}\right),\mathrm{B}\left(\overrightarrow{\mathrm{q}}\right)\text{ and }\mathrm{C}\left(\overrightarrow{\mathrm{r}}\right)$ then $\overrightarrow{\mathrm{IA}}=\overrightarrow{\mathrm{p}},\overrightarrow{\mathrm{IB}}=\overrightarrow{\mathrm{q}}\text{ and }\overrightarrow{\mathrm{IC}}=\overrightarrow{\mathrm{r}}$

Incentre I is  $\frac{\mathrm{a}\overrightarrow{\mathrm{p}}+\mathrm{b}\overrightarrow{\mathrm{q}}+\mathrm{c}\overrightarrow{\mathrm{r}}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}$  where p = BC, q=AC and r=AB 

Incentre is at the origin. 

Therefore, 

$\begin{array}{l}\frac{\mathrm{a}\overrightarrow{\mathrm{p}}+\mathrm{b}\overrightarrow{\mathrm{q}}+\mathrm{c}\overrightarrow{\mathrm{r}}}{\mathrm{a}+\mathrm{b}+\mathrm{c}}=\overrightarrow{0},\text{ or }\mathrm{a}\overrightarrow{\mathrm{p}}+\mathrm{b}\overrightarrow{\mathrm{q}}+\mathrm{c}\overrightarrow{\mathrm{r}}=\overrightarrow{0}\\ \Rightarrow \mathrm{a}\overrightarrow{\mathrm{IA}}+\mathrm{b}\overrightarrow{\mathrm{IB}}+\mathrm{c}\overrightarrow{\mathrm{IC}}=\overrightarrow{0}\end{array}$