vector analysis
Find the unit normal vector to the curve $r(t) = (3\sin t)i + (3\cos t)j + 4t k$ at point $( \pi /2, 0,1)$.
The tangent vector of this curve is $(3 \cos t)i -(3 \sin t)j + 4k$ and unit normal vector should be perpendicular to this vector at given point.
But I couldn't get the final answer. How to find unit normal vector at the given point$?$
It might be helpful to visualize what's happening here: for example, graphing the unit circle ($x^2 + y^2 = 1$) and the vector field $F$. Then consider what is happening at given points of the circle, e.g. at $(1, 0)$, and $(0, 1$.
In your comment above you observed that the vectors in $F$ go in "circle style". I think what you are observing each vector in $F$ is tangent to $C$, and tangent at some point $(x, y)$ of $C$, with each vector directed counter-clockwise.
We know that for each point $(x, y)$ that lies on $C$, the vector $n=\langle x, y\rangle$ is normal to $C$ (it's a given) at that point, and so at the point $(1, 0)$, $n$ lies along the $x$-axis, pointing in the positive $x$ direction. The vectors in $F:\;$ are each orthogonal to the corresponding vectors $n$, as you point out in your comment above. So the vector $f\in F$ is perpendicular to the x-axis at that point, and hence, must be tangent to $C$ at $(1, 0)$.
Likewise, consider what is happening at $(0, 1)$: At this point, $n$ lies along the y-axis, directed in positive $y$ direction (and normal to $C$). $f \in F$, at this point on $C$ is a "horizontal" vector, directed counter-clockwise, and hence tangent to $C$ at $(0, 1)$
You are correct that the vectors in $F = \langle -y, x \rangle$ are orthogonal to the corresponding vectors $n = \langle x, y \rangle$. We just need to know what's happening with vectors in $F$ at points of $C$, knowing $n$ is normal to $C$.
Included immediately below is a representation of vectors $n = \langle x, y\rangle$ for $(x,y) \in C$, normal to $C$:
$\quad\quad\quad\quad\quad$ 
See the image below for an representation of vectors tangent to the unit circle, representing vectors in $F$:
$\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$
Hint a:
The normal to $2x^2+3y^2-z^2=25$ at $(\sqrt7,3,4)$ is parallel to the gradient: $(4x,6y,-2z)=2(2\sqrt7,9,-4)$
The normal to $x^2+y^2-z^2=0$ at $(\sqrt7,3,4)$ is parallel to the gradient: $(2x,2y,-2z)=2(\sqrt7,3,-4)$
The common tangent to both surfaces would be perpendicular to both of these normals, which would make it parallel to their cross product $$ (2\sqrt7,9,-4)\times(\sqrt7,3,-4)=(-24,4\sqrt7,-3\sqrt7) $$
Hint b:
We are given that $2x^2+3y^2-z^2=25$ and $x^2+y^2-z^2=0$. Therefore, by subtracting the equations, we have that $$ x^2+2y^2=25 $$ which can be parametrized by $(x,y)=\left(5\cos(\theta),\frac5{\sqrt2}\sin(\theta)\right)$. For each point $(x,y)$, the $z$ coordinate can be computed from the equation from either surface.
Best Answer
Observe that your point is not on the curve, so it does not make sense to compute the unit normal at that point. Anyway, here is a general process to find the unit normal.
First of all you should parametrize your curve by arclength, which is given by \begin{align} s(t) & = \int_0^t \lVert r'(x) \rVert\,dx \\ & = 5t. \end{align} As a function of $s$, $r$ can be rewritten as $r(s) = \left(3\sin \frac{s}{5},3\cos \frac{s}{5}, \frac{4}{5}s\right)$. The unit tangent is then $$T(s) = \frac{1}{\lVert \frac{dr}{ds}\rVert}\frac{dr}{ds} = \left(\frac{3}{5}\cos \frac{s}{5}, -\frac{3}{5}\sin \frac{s}{5}, \frac{4}{5} \right).$$ The unit normal is just given by the renormalized derivative of the tangent vector: $$N(s)=\frac{T'(s)}{\lVert T'(s)\rVert} = \left(-\sin \frac{s}{5},-\cos \frac{s}{5}, 0\right). $$