The points are $A(2,3), B(4, k)$ and $C(6,-3)$

$\left(x_{1}=2, y_{1}=3\right),\left(x_{2}=4, y_{2}=k\right)$ and $\left(x_{3}=6, y_{3}=-3\right)$

Since, the points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ are collinear. Then,

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

$\Rightarrow 2(k+3)+4(-3-3)+6(3-k)=0$

$\Rightarrow 2 \mathrm{k}+6-24+18-6 \mathrm{k}=0$

$\Rightarrow-4 \mathrm{k}=0$

$\Rightarrow \mathrm{k}=0$