Plane of all points equidistant from two other points?

Question

Find the equation of the plane that contains all the points that are equidistant from the given points

$(-9, 3, 3), (6, -2, 4)$

I think the plane described lies in the midpoint of these points, and it is perpendicular to the line connecting the two points. This means that the point $\left(\dfrac{15}{2}, \dfrac{1}{2}, \dfrac{7}{2}\right)$ is on the plane, and vector line perpendicular to the plane is $<-9-6, 3-(-2), 3-4>=<-15, 5, -1>$ . So the equation of the plane is $$-15(x-\dfrac{15}{2})+5(y-\dfrac12)-(z-\dfrac72)=0$$ or $$15x-5y+z-\dfrac{227}{2}=0$$ However, this seems to be the wrong answer. What am I doing wrong?

Answer 1

The midpoint is $\left(\frac{-3}{2},\frac{1}{2},\frac{7}{2} \right)$

Answer 2

Notice, there is another easy method to find the equation of the plane

Let the parametric point be $(x, y)$ on the plane which is equidistant from the given points $(-9, 3, 3)$ & $(6, -2, 4)$ hence, we have $$\sqrt{(x-(-9))^2+(y-3)^2+(z-3)^2}=\sqrt{(x-6)^2+(y-(-2))^2+(z-4)^2}$$ $$(x+9)^2+(y-3)^2+(z-3)^2=(x-6)^2+(y+2)^2+(z-4)^2$$ $$((x+9)^2-(x-6)^2)+((y-3)^2-(y+2)^2)+((z-3)^2-(z-4)^2)=0$$ $$(x+9+x-6)(x+9-x+6)+(y-3+y+2)(y-3-y-2)+(z-3+z-4)(z-3-z+4)=0$$ $$15(2x+3)-5(2y-1)+(2z-7)=0$$ $$30x-10y+2z+43=0$$$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{\text{Equation of the plane:}\ 30x-10y+2z+43=0}}$$