After using the product rule:
$$
{\bf T}'(t)= {1\over 2e^{2t}+2e^{-2t}} \bigl< 0, 4e^{2t}, 4 e^{-2t} \bigr>
-{4e^{2t}-4e^{-2t}\over (2e^{2t}+2e^{-2t})^2 } \bigl< 2\sqrt2, 2e^{2t}, -2 e^{-2t} \bigr>
.
$$
By definition, the direction of the unit normal vector is the direction of the vector ${\bf T'}$.
To simplify things when finding the unit normal, you can multiply ${\bf T}'(t)$ by a positive scalar. This will give a vector in the same direction as that of $\bf N$; multiplying a vector by a positive number does not change its direction. (Said differently ${{\bf T}(t)\over |{\bf T}(t)|} = {|a|{\bf T}(t)\over |a{\bf T}(t)|} $
for any nonzero $a$.)
Once we have our direction vector, divide by its length to get ${\bf N}(t)$.
So, let's multiply ${\bf T}'(t)$ by $(2e^{2t}+2e^{-2t})^2/2$. This gives the vector
$$
{\bf F}(t)={ (e^{2t}+e^{-2t})} \bigl< 0, 4e^{2t}, 4 e^{-2t} \bigr>
-({2e^{2t}-2e^{-2t}}) \bigl< 2\sqrt2, 2e^{2t},- 2 e^{-2t} \bigr>
$$
which is a bit easier to deal with.
After finding $|{\bf F}(t)|$, you can compute ${\bf N}(t) ={{\bf F}(t)\over |{\bf F}(t) | }$.
(Note that when finding the curvature, you need to find $|{\bf T}'(t)|$ proper.)
There is no need to write the equation of the line , $r(t)$, which passes through $(x_0,y_0,z_0)$ and is normal to the surface. $\nabla f(x_0,y_0,z_0)$ is a vector normal to the surface $f(x,y,z)=0$ at the point $(x_0,y_0,z_0)$. So the only things you needs to do are
1) Compute the unit normal by ${\bf{n}} = {{\nabla f} \over {\left\| {\nabla f} \right\|}}$ or any other vector parallel to that which can be $ \nabla f$ itself.
2) Write the equation of the plane passing through $(x_0,y_0,z_0)$
$$ {\nabla f(\bf{x_0})} \cdot ( {\bf{x}} - {\bf{x_0}} )= \bf{0} $$
3) In your problem, computation is as follows
$$ {\bf{x_0}} = {(3,2,5)}\\
{\nabla f(\bf{x_0})}=(6,16,-10)
$$
and hence
$$\eqalign{
& 6\left( {x - 3} \right) + 16\left( {y - 2} \right) - 10\left( {z - 5} \right) = 0 \cr
& 6x + 16y - 10z + \left( { - 18 - 32 + 50} \right) = 0 \cr
& 6x + 16y - 10z = 0 \cr} $$
is the equation of the tangent plane.
Best Answer
The unit normal vector $\bf N$ is by definition ${\bf T}'\over \Vert {\bf T}'\Vert$.
Since $\bf T$ has constant norm one, it is orthogonal to $\bf T'$: $${\bf T}\cdot {\bf T}'=0$$ (if $\bf X$ has constant norm, then $0={d\over dt}({\bf X}\cdot{\bf X })= {\bf X}'\cdot{\bf X }+{\bf X}\cdot{\bf X }' =2{\bf X}' \cdot{\bf X} $)$^\dagger$.
This implies ${\bf T}\cdot {{\bf T}'\over \Vert {\bf T}'\Vert}=0$; thus $\bf N$ is normal to $\bf T$.
$^\dagger$ This is the real "reason" why $\bf N$ is orthogonal to $\bf T$. To repeat myself: if ${\bf X} $ has constant norm, then ${\bf X} (t)$ is orthogonal to ${\bf X}'(t)$. Though the product rule for differentiating a dot product is a perfectly fine way to see this, I prefer the following "proof":
Suppose ${\bf X}(t)$ has constant norm $a$ and consider it to be a position vector. Then as $t$ varies, ${\bf X}(t)$ describes a path on the surface of a sphere of radius $a$. We know that ${\bf X'}(t_0)$ is tangent to the path traced out by ${\bf X}(t)$ at the point ${\bf X}(t_0)$; so, ${\bf X'}(t_0)$ is tangent to the aforementioned sphere and, hence, orthogonal to ${\bf X}(t_0)$.