You are asking: for what values of $h$ are the vectors
$$\vec{v_1}=\left(\begin{array}{r}1\\-4\\2\end{array}\right),\quad \vec{v_2}=\left(\begin{array}{r}-3\\7\\-6\end{array}\right),\quad \vec{v_3}=\left(\begin{array}{r}4\\h\\8\end{array}\right)$$
linearly dependent?
You seem to be trying to do this by looking at the equation
$$\alpha\vec{v_1}+\beta\vec{v_2}+\gamma\vec{v_3}=\left(\begin{array}{c}0\\0\\0\end{array}\right)$$
and trying to determine for what values of $h$ there is a nonzero solution. This leads to the matrix you have:
$$\left(\begin{array}{rrr|c}
1 & -3 & 4 & 0\\
-4 & 7 & h & 0\\
2 & -6 & 8 & 0
\end{array}\right).$$
Now, since the third equation is a multiple of the first, that equation does not matter: it provides no new information. That means that you have a homogeneous system of two equations in three unknowns. Those systems always have infinitely many solutions. In particular, no matter what $h$ is, the system has infinitely many solutions, and so must have a nontrivial solution. Thus, the vectors are always linearly dependent.
To understand what is happening, note that all three vectors lie in the plane $z=2x$. Any two vectors on the plane that are not collinear will span the plane. Since $\vec{v_1}$ and $\vec{v_2}$ are not collinear, and both lie on the plane $z=2x$, any vector that lies on the plane $z=2x$ will be a linear combination of $\vec{v_1}$ and $\vec{v_2}$. Or, put another way, three vectors in a $2$-dimensional space (a plane through the origin) are always linearly dependent.
Here you have three vectors that satisfies $z=2x$; every other vector that satisfies that is a linear combination of $\vec{v_1}$ and $\vec{v_2}$: if $(a,b,2a)^t$ lies in the plane, then the system
$$\alpha\left(\begin{array}{r}1\\-4\\2\end{array}\right) + \beta\left(\begin{array}{r}-3\\7\\-6\end{array}\right) = \left(\begin{array}{c}a\\b\\2a\end{array}\right)$$
has a solution, namely $\alpha = -\frac{7a+3b}{5}$, $\beta=-\frac{4a+b}{5}$ (obtained by Gaussian elimination). In particular, since no matter what $h$ is $\vec{v_3}$ lies in the plane $2z=x$, then we will have
$$\vec{v_3} = -\frac{28+3h}{5}\vec{v_1} - \frac{16+h}{5}\vec{v_2}.$$
Note that this makes sense no matter what $h$ is.
This can be read off your row-reduced matrix: you got
$$\left(\begin{array}{rrr|c}
1 & -3 & 4 & 0\\
0 & -5 & h+16 & 0\\
0 & 0 & 0 & 0
\end{array}\right).$$
Divide the second row by $-5$ to get
$$\left(\begin{array}{rrr|c}
1 & -3 & 4 & 0\\
0 & 1 & -\frac{h+16}{5} & 0\\
0 & 0 & 0 & 0
\end{array}\right),$$
and now add three times the second row to the first row to get
$$\left(\begin{array}{rrc|c}
1 & 0 & 4+\frac{-3h-48}{5} & 0\\
0 & 1 & -\frac{h+16}{5} & 0\\
0 & 0 & 0 & 0
\end{array}\right) = \left(\begin{array}{rrc|c}
1 & 0 & -\frac{28+3h}{5} & 0\\
0 & 1 & -\frac{h+16}{5} & 0\\
0 & 0 & 0& 0
\end{array}\right).$$
So $\alpha$ and $\beta$ are leading variables, and $\gamma$ is a free variable. This tells you that the solutions to the original system are:
$$\begin{align*}
\alpha &= \frac{28+3h}{5}t\\
\beta &= \frac{h+16}{5}t\\
\gamma &= t
\end{align*}$$
Any nonzero value of $t$ gives you a nontrivial solution, and $t=-1$ gives you the solution I give above.
Of course, this can be done much more simply noting that since your original matrix has linearly dependent rows (third row is a scalar multiple of the first row), then the dimension of the rowspace is at most $2$ (in fact, exactly $2$), and hence the dimension of the columnspace is at most $2$ (in fact, exactly $2$, since $\dim(\text{columnspace})=\dim(\text{rowspace})$, so the columns are always linearly dependent.
