Said plane will have a direction vector equal to the direction between the two points: (3, 0, 0) - (-2, -4, 0) = (5, 4, 0). It will pass through the midpoint of the segment beween these points, which is ((3, 0, 0)+(-2, -4, 0))/2 = (1/2, -2, 0).
So, the equation of the plane is
... 5(x -1/2) +4(y +2) +0(z) = 0 . . . . . . plane with given direction through given point
... 5x +4y +11/2 = 0 . . . . . simplify a bit
... 10x + 8y = -11 . . . . . . . . written in standard form
