Find distance between line x-y-z =2, 2x+y+z=4 and the plane x-y=5

My problem is that the book is saying the minimum distance between the plane and line is zero but l am getting $\displaystyle 6/\sqrt{2}$. Can someone tell me where did l get this one wrong

Attempt to solution :

If l intersect the first two planes given l will get the equation of the line :

x-y-z=2................(1)

2x+y+z=4...............(2)

Add (1) and (2) ===> 3x+2z=6

z=3-3x/2.............(3)

Substitute (3) into (1)

y=1-x/2

and x=x

Equations of the line :

x=t

y=1-t/2

z=3-3t/2

Get point on the line and the normal vector for plane :

(0,1,3)

Formula for distance between point and plane x-y=5 :

normal vector for the plane : (1,-1,0)

$\displaystyle distance=\frac{ax+by+cz+d}{\sqrt{(a^2+b^2+c^2)}}$

when l solve for distance l get -6/sqrt(2).