If the points $\left(t,2t\right),\left(-2,6\right),\left(3,1\right)$  are collinear, then the value of $\text{'}t\text{'}$  is

$\Delta =\frac{1}{2}\left|t\left(6-1\right)+\left(-2\right)\left(1-2t\right)+3\left(2t-6\right)\right|$

$\frac{1}{2}\left|5t-2+4t+6t-18\right|=0$

$15t-20=0$

$t=\frac{20}{15}$

$t=\frac{4}{3}$

(or)

$\frac{6-2t}{-2-t}=\frac{1-6}{3+2}$

$\frac{2t-6}{2+t}=-1$

$2t-6=-2-t$

$3t=4$

$t=\frac{4}{3}$ .