tangent hyperplanes and tanget planes, a bit confused about dimensions

I thought I understood tangent planes until I saw tangent hyperplane and 3-d contour here. So a little bit of help will be most welcome. (i) and (ii) is no problem and agreed to the textbook answer. I am a bit stuck on (iii).

Question (Calculus, Binmore and Davies, 3.9(#5*))

Let $\displaystyle f(x,y,z)=ln(x^2+y^2+z^2)$

i. Find the gradient of f at the point $\displaystyle (x,y,z)^T$

ii. Find the tangent hyperplane to the hypersurface $\displaystyle u=ln(x^2+y^2+z^2)$

where $\displaystyle (x,y,z,u)^T=(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.

iii. Find the normal and the tangent plane to the contour

$\displaystyle ln(x^2+y^2+z^2)=0$ at $\displaystyle (1/\sqrt{3},1/\sqrt{3},1/\sqrt{3},0)^T$.

Answer.

i. $\displaystyle \nabla{f}=\frac{1}{x^2+y^2+z^2}(2x,2y,2z)^T$

ii. To find tangent hyperplane, I want to use the formula

$\displaystyle u-U=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)+(z-Z)f_z(x,y,z)$

$\displaystyle u-0=(x-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(y-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}+(z-\frac{1}{\sqrt{3}})\frac{2}{\sqrt{3}}$

$\displaystyle u=\frac{2}{\sqrt{3}}(x+y+z)-2$

iii. To me, the equation for the countour $\displaystyle ln(x^2+y^2+z^2)=0$ looks exactly the same as the equation to the surface in (ii) as u=0 in both cases.

*I understand in (ii) the hypersurface is in 4 dimentional space, and the hyperplane is in 3 dimensions. Here the contour is in 3 dimensions (??) and the tangent plane is 2-dimentional. *

*???*

I am not sure what to do, but *I was thinking may be try to express z as the function of x and y and apply the same template as in (ii) ie*

*$\displaystyle z-Z=(x-X)f_x(x,y,z)+(y-Y)f_y(x,y,z)$*

*???*