Answer
$\langle 6,2,-4\rangle$
Work Step by Step
Given $$x^{2}+y^{2}-z^{2}=6,\ \ \ \ P=( 3,1,2)$$
Consider $f(x,y,z)= x^{2}+y^{2}-z^{2}-6$
Since
\begin{align*}
\nabla f&=\left\langle\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\right\rangle\\
&=\left\langle 2x,2y,-2z\right\rangle\\
\nabla f\bigg|_{(3,1,2)}&= \left\langle 6,2,-4\right\rangle
\end{align*}
Then the normal vector on the surface $x^{2}+y^{2}-z^{2}=6$ at $P$ is $\langle 6,2,-4\rangle$
