Find the equation of the plane that passes through the point $(8,-6,-7)$ and is parallel to the plane $9x-y-z= 8$.

Using the linear definition:

$9x-y-z-8=0$ is a plane with normal vector (9,-1,-1). So I think I can use this as my normal vector for the plane since two parallel planes would have the same normal vector?

Correct. And shift coordinates to your new point.

$9 (x-8) - (y+6) - (z+7) = 0$ gives you the new plane

Alternatively, a plane is parallel to $9x-y-z = 8$ if and only if it is $9x-y-z = k$ for some constant $k$. We can find this specific $k$ by plugging in the point $(8, -6, -7)$:

$$ k = 9(8) - (-6) - (-7) = 85 $$

So the equation of the plane is $9x-y-z = 85$. Note that this isn't any different *in essence* from Phillip's and your solution (how could it be, really?); it's just a slightly different way of looking at the same problem.

your new plane will have an equation of the form

$9x-y-z=D$

the point $(8,-6,-7)$ is in the plane if

$72+6+7=D=85$

thus the equation we are looking for is

$9x-y-z=85$.