vert plane: x - root(3)*y + 2*root(3) - 1 = 0
surface: z = x^2 + y^2
slope of the< tangent line to the curve of intersection of the vertical plane &surface at point (1,2,5)
First, I took the gradients and plugged in (1,2,5)
del F (1,2,5) = <1, - root 3, 0> && del G (1,2,5) = <2,4,-1>
Then take cross product, gives < root 3, 1, 4+2*root(3)>. The cross product is tangent to the curve of intersection.
Now in order to get slope, I think I have to do a directional derivative, but not sure.