The points A(3,1), B(12, – 2)

True

Let $\quad A \equiv\left(x_{1}, y_{1}\right) \equiv(3,1), B \equiv\left(x_{2}, y_{2}\right) \equiv(12,-2)$

and $\quad C=\left(x_{3}, y_{3}\right)=(0,2)$

$\therefore \quad$ Area of $\triangle A B C \Delta=\frac{1}{2}\left[x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)\right]$

$=\frac{1}{2}[3-(2-2)+12(2-1)+0\{1-(-2)\}]$

$=\frac{1}{3}[3(-4)+12(1)+0]$

$=\frac{1}{3}(-12+12)=0$

$\because \quad$ Area of $\triangle A B C=0$

Hence, the points $A(3,1), B(12,-2)$ and $C(0,2)$ are collinear.

So, the points $A(3,1), B(12,-2)$ and $C(0,2)$ cannot be the vertices of a triangle.