If the points $(k, 2k),\ ( 3k, 3k)$ and $(3, 1)$ are collinear then the value of $k$ is

Since the given points are collinear, they do not form a triangle, which means area of the triangle is Zero

The area of the triangle with vertices $({ x }_{ 1 },{ y }_{ 1 })$ ; $({ x }_{ 2 },{ y}_{ 2 })$ and $({ x }_{ 3 },{ y }_{ 3 })$ is $\left| \frac { {x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{3 }({ y }_{ 1 }-{ y }_{ 2 }) }{ 2 } \right|$

Hence, substituting the points $({ x }_{ 1 },{ y }_{ 1 }) = (k,2k)$ ; $({ x}_{ 2 },{ y }_{ 2 }) = (3k,3k)$ and $({ x }_{ 3 },{ y }_{ 3 }) = (3,1)$ in the area formula, we get $\left| \frac { k(3k-1)+3k(1-2k)+3(2k-3k) }{ 2 } \right| =0$

$=> 3{k}^{2} -k + 3k-6{k}^{2} + 6k -9k = 0$

$=>-3{k}^{2} -k = 0$

$-k(3k+1) =0$

$k = 0$ or $k = -\frac{1}{3}$