#### Answer

The system is consistent for all $h$.

#### Work Step by Step

We begin with matrix
$\begin{bmatrix}
1 & h & -3 \\
-2 & 4 & 6 \\
\end{bmatrix}$
First, we need to multiply the first row by $2$ and add it to the second row:
$\begin{bmatrix}
1 & h & -3 \\
0 & 2h+4 & 0 \\
\end{bmatrix}$
Write $c$ for $2h+4$. Then the second equation $c\cdot x_2 = 0$ has a solution for every value of $c$. So the system is consistent for all $h$.