Find the equation of the plane passing through the intersection of $\displaystyle 2x+y+2z=9$ and $\displaystyle 4x-5y-4z=1$ and passing through $\displaystyle (3,2,-1)$

I forgot how these sums are done:Here's what I did:

solving the two equations we get $\displaystyle x=\frac{23-3z}{7}$ and $\displaystyle y=\frac{17-8z}{7}$

hence we get the parametric equation:$\displaystyle <\frac{23-3t}{7},\frac{17-8t}{7},t>$

hence the direction vectors=$\displaystyle <\frac{-3}{7},\frac{-8}{7},1>$

as the line is perpendicular to the intersection, it is also the normal vector of the 3rd plane:

equation of required plane:$\displaystyle -\frac{3}{7}x-\frac{8}{7}y+z=D$

putting $\displaystyle (x,y,z)=(3,2,-1)$ we get $\displaystyle D=\frac{32}{7}$

new simplified equation:$\displaystyle 3x+8y-7z=32$

but the answer given in the book is: $\displaystyle 11x-5y-z=24$

What am I doing wrong here?

thanks in advance.