Let's say our surface is and we want to find tangent plane and the parametric equations of the normal line in (1,2,1). So .
1) Gradient of F is (6x-y, -x, -1), grad F(1,2,1) = (4,-1,-1). So tangent plane will have the equation that is a scalar multiplication of grad F(1,2,1) on (x-1,y-2,z-1) because (1,2,1) is a point in which we're looking for tangent plane, so 4(x-1)-(y-2)-(z-1)=0, which is 4x-y-z-1=0. Answer is 4x-y-z=1.
2) Each of the equations x(t), y(t), z(t) will be the sum of x, y, z component of a point (1,2,1) and corresponding of vector grad F(1,2,1) multiplied by parameter t. So if grad F(1,2,1)=(4,-1,-1) and point (1,2,1) then the parametric equations of the normal line will look like this:
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