If $\mathrm{p},\mathrm{q}$ are two non collinear and non zero vectors such that
$\left(\mathrm{b}-\mathrm{c}\right)\left(\overline{\mathrm{p}}\times \overline{\mathrm{q}}\right)+\left(\mathrm{c}-\mathrm{a}\right)\overline{\mathrm{p}}+\left(\mathrm{a}-\mathrm{b}\right)\overline{\mathrm{q}}=\overline{0}$ where $\mathrm{a},\mathrm{b},\mathrm{c}$ are the lengths of the sides of a triangle, then the triangle is 
 

$$\begin{gathered} \text{ Given }p\text{ and }q\text{ are non collinear and non zero vectors } \\ (b-c)(p\times q)+(c-a)p+(a-b)q=0 \\ \text{ Since }\mathrm{p}\text{ and }q\text{ are non zero and non collinear vectors },\mathrm{p}\times \mathrm{q}\text{ is also a non zero vector} \\ \text{ Therefore for the given equation to be true }\mathrm{b}-\mathrm{c}=0,\mathrm{c}-\mathrm{a}=0\text{ and }a-b=0 \end{gathered}$$

its possible for b=c=a

 $\therefore ∆\mathrm{le}$  equilateral