What is the Normal Vector to the Surface S at the Points of Intersection?

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Let S be the surface given by the equation 9x^2 + y^2 − z^2 − 2y + 2z = 1, Show that the straight line r(t) = <1, 1, 1> + t<1, 0, 0> is normal to the surface S at the points of intersection.

I set both equations equal to each other and I found their points of intersection are (1/3,1,1) and (-1/3,1,1). But I don't know where to go from there. :confused:
 
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Compute the derivative of the equ. of the surface with respect to the variables. Those are the tengential directions, the cross product is hence giving the normal at a given point...then it should be easy
 
Or compute the tangent planes at the points of intersection and show the line is perpendicular to the planes.
 
the normal vector to a surface f= 0 is the gradient of f at the given point.
 

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