(B) False.
Let the points be $A(0, 5)$,$B(0, -9)$,and $C(3, 6)$.
For points to be collinear,the area of the triangle formed by them must be zero.
The area of a triangle with vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$,and $(x_{3}, y_{3})$ is given by:
$\Delta = \frac{1}{2} |x_{1}(y_{2} - y_{3}) + x_{2}(y_{3} - y_{1}) + x_{3}(y_{1} - y_{2})|$
Substituting the values $x_{1}=0, y_{1}=5, x_{2}=0, y_{2}=-9, x_{3}=3, y_{3}=6$:
$\Delta = \frac{1}{2} |0(-9 - 6) + 0(6 - 5) + 3(5 - (-9))|$
$\Delta = \frac{1}{2} |0 + 0 + 3(5 + 9)|$
$\Delta = \frac{1}{2} |3 \times 14| = \frac{42}{2} = 21$
Since the area of the triangle is $21 \neq 0$,the points are not collinear.