Answer
a. There are no values of $h$ such that $\vec{v}_{3}$ is in $Span\{\vec{v}_{1},\vec{v}_{2}\}$.
b. The set is linearly dependent for all values of $h$.
Work Step by Step
We can prove both parts with a single line of reasoning. By row reduction, we see that $\begin{bmatrix}1&-3&5\\-3&9&-7\\2&-6&h\end{bmatrix}\sim\begin{bmatrix}1&-3&5\\0&0&8\\0&0&h-10\end{bmatrix}$.
That the second column contains no pivot indicates 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 $(0,0,8)$. Hence, no value of $h$ puts $\vec{v}_{3}$ in $Span\{\vec{v}_{1},\vec{v}_{2}\}$.
