If the line passing though the points (5, 1, a) and (3, b, 1) crosses the yz plane at the point $\left(0,\frac{17}{2},-\frac{13}{2}\right)$, then 

The equation of the line passing through (5, 1, a) and (3, b, 1) is $\frac{\mathrm{x}-5}{-2}=\frac{\mathrm{y}-1}{\mathrm{b}-1}=\frac{\mathrm{z}-\mathrm{a}}{1-\mathrm{a}}$
It cuts yz-plane at $\left(0,\frac{17}{2},\frac{-13}{2}\right)$
This means that this point lies on the line $\therefore \frac{5}{2}=\frac{\frac{17}{2}-1}{\mathrm{b}-1}=\frac{\frac{-13}{2}-\mathrm{a}}{1-\mathrm{a}}$
$\Rightarrow \mathrm{b}-1=3\text{ and }-13-2\mathrm{a}=5(1-\mathrm{a})\Rightarrow \mathrm{b}=4\text{ and }\mathrm{a}=6$