For what value of x are the points A

 A(−3, 12), B(7, 6) and C(x, 9) are the given points. Then:
(x1 = −3, y1 = 12), (x2 = 7, y2 = 6) and (x3 = x, y3 = 9)
It is given that the points A, B and C are collinear. Therefore,

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

$\Rightarrow(-3)(6-9)+7(9-12)+x(12-6)=0$

$\Rightarrow(-3)(-3)+7(-3)+x(6)=0$

$\Rightarrow 9-21+6 x=0$

$\Rightarrow 6 x-12=0$

$\Rightarrow 6 x=12$

$\Rightarrow x=\frac{12}{6}=2$

Therefore, when x= 2, the given points are collinear.