If $\overrightarrow{x}\text{ and }\overrightarrow{y}$ are two non-collinear vectors and $ABC$is a triangle with sides $a, b, c$ satisfying 

$(20a-15b)\overrightarrow{x}+(15b-12c)\overrightarrow{y}+(12c-20\overrightarrow{a})(\overrightarrow{x}\times \overrightarrow{y})=\overrightarrow{0}$ 

then the triangle $ABC$  is

Vectors $\overrightarrow{x},\overrightarrow{y}\text{ and }\overrightarrow{x}\times \overrightarrow{y}$  are non-coplanar 

vectors.

$\begin{array}{l}\therefore (20a-15b)\overrightarrow{x}+(15b-12c)\overrightarrow{y}+(12c-20a)(\overrightarrow{x}\times \overrightarrow{y})=\overrightarrow{0}\\ \Rightarrow 20a-15b=0,15b-12c=0\text{ and }12c-20a=0\\ \Rightarrow 4a=3b,5b=4c\text{ and }3c=5a\\ \Rightarrow \frac{a}{3}=\frac{b}{4}=\frac{c}{5}\mid =\left(\lambda \text{ say }\right)\Rightarrow a=3\lambda ,b=4\lambda ,c=5\lambda \end{array}$

Clearly, $c^2=a^2+b^2$

Hence, $∆ABC$ is a right angled triangle.