Answer
a. No value of $h$ sets $\vec{v}_{3}$ in $Span\{\vec{v}_{1},\vec{v}_{2}\}$.
b. All values of $h$ make the vectors linearly dependent.
Work Step by Step
We can prove both parts by a single line of reasoning. Row-reducing the matrix formed from the column vectors, we get that
$\begin{bmatrix}1&-2&2\\-5&10&-9\\-3&6&h\end{bmatrix}\sim\begin{bmatrix}1&-2&2\\0&0&1\\0&0&6+h\end{bmatrix}$.
We can see from the first and second columns that $\mathbf{A}\vec{x}=\vec{0}$ has infinitely many solutions, and so does the corresponding vector equation. This means that the vectors are linearly dependent for all values of $h$. However, it is also clear that the third column cannot be formed from a sum of the first two columns, owing to the row $(0,0,1)$. Hence, no value of $h$ puts $\vec{v}_{3}$ in $Span\{\vec{v}_{1},\vec{v}_{2}\}$.
