Let $a$ be the answer. By taking the $z$-axis to be the cross product of these two lines, we can let the lines be on the planes $z=0$ and $z=h$, respectively. Then, by projecting onto the $xy$-plane, we get the above diagram. The projected lengths of the first four segments are $\sqrt{25-h^{2}}, \sqrt{9-h^{2}}$, and $\sqrt{4-h^{2}}$, and $\sqrt{a^{2}-h^{2}}$. By similar triangles, these lengths must form a geometric progression. Therefore, $25-h^{2}$, $9-h^{2}, 4-h^{2}, a^{2}-h^{2}$ is a geometric progression. By taking consecutive differences, $16,5,4-a^{2}$ is a geometric progression. Hence, $4-a^{2}=\frac{25}{16} \Longrightarrow a=\frac{\sqrt{39}}{4}$.