Find a unit tangent vector to a curve that is an intersection of two surfaces. The intersection of the two surfaces given by the Cartesian equations $2x^2+3y^2-z^2=25$ and $x^2+y^2=z^2$ contains a curve $C$ passing through the point $P=(\sqrt{7},3,4)$. These equations may be solved for $x$ and $y$ in terms of $z$ to give a parametric representation of $C$ with $z$ as a parameter.
(a) Find a unit tangent vector $T$ to $C$ at the point $P$ without using an explicit knowledge of the parametric representation.
(b) Check the result in part (a) by determining a parametric representation of $C$ with $z$ as a parameter.
For (b), I solved the two equations for the surfaces given to get $y^2=25-z^2$ and $x^2=2z^2-25$. Since we're looking for the curve that contains $P$, $x$, $y$ should be positive so we get $y=\sqrt{25-z^2}$ and $x=\sqrt{2z^2-25}$. So from this I get the parametric representation $(\sqrt{2z^2-25}, \sqrt{25-z^2},z)$ for the curve $C$. 
Is this the correct way of finding the parametrization? Moreover, I do not know how to find the unit targent vector $T$ to $C$ at $P$, without getting a parametrization. How can I find this? The answer to $T$ is $\frac{1}{\sqrt{751}}(24,-4\sqrt{7},3\sqrt{7})$. 
I would greatly appreciate any solutions, hints or suggestions.

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.
