The cross section of a right prism is an equilateral triangle. The rectangular face ABCD of the prism lies on the plane $\displaystyle x+y+5z=\sqrt{2}$, where A and B are the points $\displaystyle (0,\sqrt{2},0)$ and $\displaystyle (\sqrt{2},0,0)$ respectively. EF is the edge in which the other two rectangular faces ADEF and BCEF meet.

Prove that the equation of the plane containing A, B, F is

$\displaystyle 5x+5y-2z=5\sqrt{2}$.

If the origin lies inside the prism, determine the equations of the line EF.

I have proven the first part, but I can't get the equations of EF.

F(x,y,z)

AF=BF from this x=y

AB=AF=BF=2

Perpendicular distance of F from plane ABCD=$\displaystyle \sqrt 3=AF(\cos \theta)$

Where $\displaystyle \theta$ is the angle of elevation of F from A.

$\displaystyle \cos\theta=[(\frac{1}{3\sqrt 3})(\frac{x-\sqrt 2}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)})+(\frac{1}{3\sqrt 3})(\frac{y}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)})+(\frac{5}{3\sqrt 3})(\frac{z}{\surd(x^2-2\sqrt 2x+2+y^2+z^2)})]$

Fill in the values for the perpendicular distance.

Then we have $\displaystyle 9=\frac{2x-2\sqrt 2+2y+10z}{\surd (x^2-2\sqrt 2x+2+y^2+z^2)}$

BF=$\displaystyle \surd (x^2-2\sqrt 2x+2+y^2+z^2)=2$

So

$\displaystyle 9=\frac{2x-2\sqrt 2+2y+10z}{2}$

$\displaystyle 9=x-\sqrt 2+y+5z$

$\displaystyle 5x+5y=45+5\sqrt 2-25z=5\sqrt 2+2z$ from the plane ABF

$\displaystyle z=\frac{5}{3}$

now x=y

$\displaystyle 2x=9+\sqrt 2-5(\frac{5}{3})$

$\displaystyle x=y=\frac{1}{3}+\frac{\sqrt 2}{2}$

So $\displaystyle F(\frac{1}{3}+\frac{\sqrt 2}{2}, \frac{1}{3}+\frac{\sqrt 2}{2}, \frac{5}{3})$

The direction ratios of the line are 5:5:-2 since it is perpendicular to the plane ABF

the equations for the line I arrive at are

$\displaystyle -2x=-2y=5z-9-\sqrt 2$

the answer is supposed to be $\displaystyle -2x=-2y=5z+9-\sqrt 2$

I can't find where I'm wrong.

Thanks!