Assuming you wrote
[tex]a_1=\begin{bmatrix}-1&h&7\end{bmatrix}^\top[/tex]
[tex]a_2=\begin{bmatrix}4&-2&5\end{bmatrix}^\top[/tex]
[tex]a_3=\begin{bmatrix}1&-7&2\end{bmatrix}^\top[/tex]
[tex]a_1,a_2,a_3[/tex] will span [tex]\mathbb R^3[/tex] provided that they comprise a set of linearly independent vectors. This happens when the only solution to the following is [tex]\mathbf c=\mathbf0[/tex]:
[tex]\begin{bmatrix}a_1&a_2&a_3\end{bmatrix}\mathbf c=\mathbf0[/tex]
[tex]\iff\underbrace{\begin{bmatrix}-1&4&1\\h&-2&-7\\7&5&2\end{bmatrix}}_{\mathbf A}\begin{bmatrix}c_1\\c_2\\c_3\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}[/tex]
There will be a non-zero solution whenever [tex]\mathbf A[/tex] is singular, i.e. when [tex]\det\mathbf A=0[/tex]. You have
[tex]\det\mathbf A=\begin{vmatrix}-1&4&1\\h&-2&-7\\7&5&2\end{vmatrix}=-3h-213=0[/tex]
which has a solution of [tex]h=-71[/tex]. This is the only value that makes the set of vectors linearly dependent, which means all [tex]h\neq-71[/tex] guarantee that [tex]a_1,a_2,a_3[/tex] span all of [tex]\mathbb R^3[/tex].
