Reducing the System of linear equations \begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 \\
4x+y+(k^2-14)z&=k+2
\end{align*}
I started doing the matrix of the system:
$$ \begin{pmatrix} 1 & 2 & -3 & 4 \\ 3 & -1 & 5 & 2 \\ 4 & 1 & k^2-14 & k+2  \end{pmatrix} $$
But I still don't know how can I reduce this, could you explain me what steps I should follow?
Which values of $k$ make the system inconsistent, determined, undetermined?

This is not very different from other answers. And I think that the approach using Gaussian elimination is more systematic. I wanted to show how this can be done with a little of guesswork.

If you notice that the sum of the first two equations is $4x+y+2z=6$, then you see that every solution of your system also fulfills the equations
\begin{align*}
4x+y+2z&=6 \\
4x+y+(k^2-14)z&=k+2
\end{align*}
Now if you subtract these equations you get:
$$(k^2-16)z=k-4$$
which is the same as $(k-4)(k+4)z=(k-4)$.
If $\boxed{k=4}$, then this last equation is $0=0$ and you system is equivalent to the system
\begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 
\end{align*}
If $\boxed{k=-4}$, then the this equation is $0=-8$, and the system does not have solution.
If $\boxed{k\ne\pm4}$, then the original system is equivalent to the following system of equations:
\begin{align*}
x+2y-3z&=4 \\
3x-y+5z&=2 \\
(k^2-16)z&=k-4
\end{align*}
Since $k^2-16\ne0$, the variable $z$ is uniquely determined by the last equation. Since the matrix $\begin{pmatrix}1&2\\3&-1\end{pmatrix}$ is regular, the values of $x$ and $y$ are the also uniquely determined.
