True
Let A=(x1,y1)=(3,1),B=(x2,y2)=(12,−2)
and C=(x3,y3)=(0,2)
∵Area of ΔABC=12[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]=12[3(−2−2)+12(2−1)+0(1−(−2))]=13[3(−4)+12(1)+0]=13(−12+12)=0∵Area of ΔABC=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.