The value of k for which the points $A(2,3),B(4,k),$ and $C(6,-3)$ are collinear.

The correct option is C: $k=0$

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

Here, $x_1=2,y_1=3,x_2=4,y_2=k,x_3=6,y_3=-3$

When the points are collinear, the area of the triangle will be zero.

Area of the triangle $=\frac{1}{2}\left[x_1\right(y_2-y_1)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]$

The points are collinear.

$\therefore \frac{1}{2}\left[x_1\right(y_2-y_1)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$

$\Rightarrow \frac{1}{2}\left[2\right(k-3)+2(-3-3)+6(3-k\left)\right]=0$

$\Rightarrow \frac{1}{2}[2k-6-12+18-6k]=0$

$\Rightarrow |-2k|=0$ [Area cannot be negative]

$\Rightarrow 2k=0$

$\Rightarrow k=0$