If the points A(2, 3) B(4, k) and C(6, −3) are collinear, find the value of k.

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

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

It is given that the points $A, B$ and $C$ are collinear. Then,

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

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

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

$\Rightarrow-4 k=0$

$\Rightarrow k=0$