Finding the directional derivative parallel to an intersection of planes We must find the directional derivative of $ f(x,y,z) = x^2 + 2xyz -y^2 $ at $ (1,1,2) $ in a direction parallel to the straight line $ \frac{x-1}{2} = y-1 = \frac{z-2}{-3} $ 
The straight line is an intersection of 2 planes but in 3 variables, how do I write this in the form of a vector so I can find the directional derivative?

That equation is not an intersection of planes, but just the standard form of a line in 3D space. You can see by setting
$$ \frac{x-1}{2} = y - 1 = \frac{z-2}{-3} = t $$ 
Solving for $x$, $y$, $z$ each in terms of $t$ yield
$$ x = 1 + 2t $$
$$ y = 1 + t $$
$$ z = 2 - 3t $$
which is, of course, the parametric form
