If points (t, 2t), (−2, 6) and (3, 1)

We have three collinear points $\mathrm{A}(t, 2 t) ; \mathrm{B}(-2,6) ; \mathrm{C}(3,1)$.

In general if $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ are collinear then,

$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

So,

$t(6-1)-2(1-2 t)+3(2 t-6)=0$

So,

$5 t+4 t+6 t-2-18=0$

So,

$15 t=20$

Therefore,

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

So the answer is (b)