Let [tex]F(x,y,z)=f(x,y)-z[/tex]. The tangent plane to [tex]f(x,y)[/tex] at (1, 2, -6) has equation
[tex]\nabla F(1,2,-6)\cdot(x-1,y-2,z+6)=0[/tex]
We have
[tex]\nabla F(x,y,z)=(-4x,-6y,-1)\implies\nabla F(1,2,-6)=(-4,-12,-1)[/tex]
Then the tangent plane has equation
[tex](-4,-12,-1)\cdot(x-1,y-2,z+6)=0\implies -4(x-1)-12(y-2)-(z+6)=0\implies 4x+12y+z=22[/tex]
Let [tex]g(x,y)=4-x^2-y^2[/tex], and [tex]G(x,y,z)=g(x,y)-z[/tex]. The tangent plane to [tex]S[/tex] at a point [tex](a,b,c)[/tex] is
[tex]\nabla G(a,b,c)\cdot(x-a,y-b,z-c)=0[/tex]
We have
[tex]\nabla G(x,y,z)=(-2x,-2y,-1)\implies \nabla G(a,b,c)=(-2a,-2b,-1)[/tex]
so that this plane has equation
[tex](-2a,-2b,-1)\cdot(x-a,y-b,z-c)=0\implies2ax+2by+z=2a^2+2b^2+c[/tex]
In order for this plane to be parallel to the previous plane, we need to have
[tex]\begin{cases}2a=4\\2b=12\end{cases}\implies a=2,b=6\implies g(a,b)=c=-36[/tex]
so the point we're looking for is (2, 6, -36).
