The equation of the circle which passes through the points $(1,0),(0,1)$ and $(0,0)$ is
$x^{2}+y^{2}-x-y=0$
Given that, the point $(2 k, 3 k)$ is on the circle and form concyclic circle. Then, it satisfies the Eq. (i)
$(2 k)^{2}+(3 k)^{2}-(2 k)-(3 k)=0$
$\Rightarrow 4 k^{2}+9 k^{2}-5 k=0$
$\Rightarrow 13 k^{2}-5 k=0$
$\Rightarrow k(13 k-5)=0$
$\Rightarrow k=0$ or $k=\frac{5}{13}$
Hence, $k=\frac{5}{13}$