If the line passing through 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 

$a= 6, b = 4 $

The equation of the line passing through (5, 1, a) and (3, b, 1) is

$\frac{x-5}{-2}=\frac{y-1}{b-1}=\frac{z-a}{1-a}$

The coordinates of any point on this line are

$(-2r + 5, (b-1) r+1, (1-a) r + a)$

If it lies on yz-plane. Then,

$-2r + 5 = 0 ⇒ r=\frac{5}{2}$

Thus, the line cuts yz-plane at $\left(0, \frac{5b-3}{2}, \frac{5-3a}{2}\right)$

But, the coordinates of this point are given as $\left(0, \frac{17}{2}, \frac{-13}{2}\right)$

$∴ \frac{5b-3}{2}=\frac{17}{2}$ and $\frac{5-3a}{2}=\frac{-13}{2}⇒ b = 4 $ and $ a = 6 $