Analytic Geometry question, planes and lines Let there be a plane going through three points $(0,2,-9), (0,-1,0), (-\frac{3}{m},1,-3)$.
For which value of $m$ is the line $l: (3,0,-9)+t(2m,-5,7)$ onto (or 'inside') the plane? 
Not sure how to do this. Thanks in advance! 

We show m = 2. Let A = (0,2,-9), B = (0,-1,0), and C = (-3/m,1, -3). Then vectors AB = (0,-3,9), and AC = (-3/m,-1,6). The normal to the plane is N  = AB * AC = (-9,-27/m,-9/m). So the equation of the plane is: -9(x-0) -27/m*(y-2) -9/m(z+9) = 0. The line l is on the plane if two of its points are on it. So let t = 0 we get the point (3,0,-9) on it and this means its coordinates satisfies the equation of the plane and this gives: m = 2. Put t = 1 get the second point (3+2m,-5,-2) and plug the coordinates of this point into the plane equation yields:  3 + 2m  = 14/m. So m = 2.
