For what values of h are the three vectors linearly dependent? 
For the vectors below find the value(s) of h for which the vectors are linearly dependent.
$$
\vec v_1=\left[\begin{array}{c} 2 \\ -4 \\ 1 \end{array} \right]\ \vec v_2 = \left[\begin{array}{c} -6 \\ 7 \\ -3 \end{array} \right]\ \vec v_3 = \left[\begin{array}{c} 8 \\ h \\ 4 \end{array} \right]
$$

Attempt:
I started by reducing the matrix into echelon form but am confused on where to go from here and how to tell what values of h would make the vectors linearly dependent. Any help is greatly appreciated.
$$
\begin{bmatrix} 2 & -6 & 8 \\ -4 & 7 & h \\ 1 & -3 & 4 \end{bmatrix} \sim> \begin{bmatrix} 1 & 0 & \frac{-3h}{5} - \frac{28}{5} \\ 0 & 1 & \frac{-h}{5} - \frac{16}{5} \\ 0 & 0 & 0 \end{bmatrix}
$$
EDIT: Someone asked for me to show the work I did to get from the LHS to RHS. It is below:

*

*$R_1 = \frac{R_1}{2}$
$$
\begin{bmatrix} 1 & -3 & 4 \\ -4 & 7 & h \\ 1 & -3 & 4 \end{bmatrix}
$$

*$R_2 = R_2 + 4(R_1)$
$$
\begin{bmatrix} 1 & -3 & 4 \\ 0 & -5 & h+16 \\ 1 & -3 & 4 \end{bmatrix}
$$

*$R_3 = R_3 - R_1$
$$
\begin{bmatrix} 1 & -3 & 4 \\ 0 & -5 & h+16 \\ 0 & 0 & 0 \end{bmatrix}
$$

*$R_2 = -\frac{R_2}{5}$
$$
\begin{bmatrix} 1 & -3 & 4 \\ 0 & 1 & -\frac{h}{5} - \frac{16}{5} \\ 0 & 0 & 0 \end{bmatrix}
$$

*$R_1 = R_1 + 3(R_2)$
$$
\begin{bmatrix} 1 & 0 & \frac{-3h}{5} - \frac{28}{5} \\ 0 & 1 & \frac{-h}{5} - \frac{16}{5} \\ 0 & 0 & 0 \end{bmatrix}
$$
 A: Your attempt it is correct, we only need to interpret in terms of the definitions what is happening.  We always can use the definition: we says that $\{v_1,v_2,v_3\}$ is linearly dependent  if there exists scalars $\alpha_1,\alpha_2,\alpha_3$ not all zeros such that $\alpha_{1}v_1+\alpha_{2}v_2+\alpha_{3}v_3=0$. In terms the linear system equation in matrix form we have $$\begin{bmatrix}2&-6&8\\-4&7&h\\1&-3&4 \end{bmatrix}\begin{bmatrix}\alpha_1\\\alpha_2\\\alpha_3\end{bmatrix}=\begin{bmatrix}0\\0\\0 \end{bmatrix}\iff \left[\begin{array}{ccc|c}2&-6&8&0\\-4&7&h&0\\1&-3&4&0 \end{array}\right]$$
But the linear system is homogeneous (i.e., $Ax=0$) so we know that always it is consistent. Moreover if it has only solution we have independence linear but with infinite solutions we have dependence linear. We can also notice that if at least one row after of the elimination to become in zero, then we have dependence linear. So we only need look at the matrix $$\begin{bmatrix}2&-6&8\\-4&7&h\\1&-3&4 \end{bmatrix}$$ and then to make elimination by row to see under what situation a row could become the zero row, which is our study objective. Here comes your work, let's try to clarify this.
\begin{align*}
\begin{bmatrix}2&-6&8\\-4&7&h\\1&-3&4 \end{bmatrix}&\sim \begin{bmatrix}1&-3&4\\-4&7&h\\2&-6&8 \end{bmatrix}\\
&\sim \begin{bmatrix}1&-3&4\\0&-5&h+16\\2&-6&8\end{bmatrix}\\
&\sim \begin{bmatrix}1&-3&4\\0&1&\frac{-h-16}{5}\\0&0&0 \end{bmatrix}
\end{align*}
Now, you can see (or verify) that in no step of the row reduction was a condition/constraint placed on the value of $h$. But we have obtained a row full of zeros, so the set can never be independent set regardless of the value of $h$, that is basically Anne Bauval's solution in the comments. Therefore we conclude that for all value $h$ the set $\{v_1,v_2,v_3\}$ is always dependent.
I think also as it was suggested in the comments that the way using the idea of this concept in terms of determinants can make the work easier. But it looks like you haven't seen the idea yet, so it can't be used yet.
