True

Let the coordinates of A = (x₁, y₁) = (3, 1)

Coordinates of B = (x₂, y₂) = (12, - 2)

Coordinates of C = (x₃, y₃) = (0, 2)

Area of ∆ABC = ∆ = 1/2 [x₁ (y₂ - y₃ ) + x₂ (y₃ - y₁ ) + x₃ (y₁ - y₂ )]

Δ= 1/2 [3 - (2 - 2) + 12(2 - 1) + 0{1 - ( - 2)}]

Δ =1/2 [3( - 4) + 12(1) + 0]

Δ =1/2 ( - 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) can’t be the vertices of a triangle.