Find the correct option regarding given points $(1, 2), (2, 4)$ and $(3, 6)$

Area of the triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is:

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

Here the points are $(1,2)$, $(2,4)$ and $(3,6)$, therefore, the area of the triangle is:

$A=\dfrac { 1 }{ 2 } \left| x_{ 1 }(y_{ 2 }-y_{ 3 })+x_{ 2 }(y_{ 3 }-y_{ 1 })+x_{ 3 }(y_{ 1 }-y_{ 2 }) \right| \\ A=\dfrac { 1 }{ 2 } \left| 1(4-6)+2(6-2)+3(2-4) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| 1(-2)+2(4)+3(-2) \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -2+8-6 \right| \\ \Rightarrow A=\dfrac { 1 }{ 2 } \left| -8+8 \right| \\ \Rightarrow A=0$

Since the area of the triangle is zero,

Hence, the points are collinear.