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.
We want to find all (only) those value(s) that will make the vectors linearly dependent.
Can you see, for example, why $\,a = -\frac 12\,$ is a problem? Why would $\bf \,a = -\frac 12\,$ make the vectors linearly dependent? And why would $\bf\, a = 1\,$ make the vectors linearly dependent?
A set of vectors, in your case, in $\mathbb R^3$, is linearly dependent if any one of them can be written as a linear combination of the others. In either of the above cases, $\,a = -\frac 12, \,\text{ or}\; a = 1,\,$ one or more of the vectors can be expressed as a linear combination of the others.
Recall: The determinant of an $n\times n$ matrix equals zero $\iff$ (when and only when) its column vectors are linearly dependent.
So you can also solve for $a$ by setting up the matrix using your vectors as columns in a $3 \times 3$ matrix; find the determinant, which will be a function of $a$, set it equal to zero, and solve for the $a$ values that make the determinant equal to zero (find the zeros of the determinant). Those and only those values are values for which the vectors are linearly dependent.
Best Answer
That reduced matrix shows you that the set of vectors is linearly dependent for every value of $h$. If $h\ne 10$, the system has no solution, and if $h=10$, it has infinitely many, so there is no value of $h$ that gives it exactly one solution.
Indeed, you can see this directly from the vectors themselves: $v_2=-3v_1$.