If the points (k, 2k), (3k, 3k) and (3, 1) are collinear, then k

We have three collinear points $\mathrm{A}(k, 2 k) ; \mathrm{B}(3 k, 3 k) ; \mathrm{C}(3,1)$.

In general if $\mathrm{A}\left(x_{1}, y_{1}\right) ; \mathrm{B}\left(x_{2}, y_{2}\right) ; \mathrm{C}\left(x_{3}, y_{3}\right)$ are collinear then, area of the triangle is 0 .

$x_{1}\left(y_{2}-y_{3}\right)+x_{2}\left(y_{3}-y_{1}\right)+x_{3}\left(y_{1}-y_{2}\right)=0$

So,

$k(3 k-1)+3 k(1-2 k)+3(2 k-3 k)=0$

So,

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

Take out the common terms,

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

Therefore,

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

So the answer is (b)