Let the two given points be $P(5,1,a), \quad Q(3,b,1).$

The line through $P$ and $Q$ can be written in parametric form as $\vec r = (5,1,a) + t\big((3,b,1)-(5,1,a)\big).$ So the direction vector is $(-2,\, b-1,\, 1-a).$

Hence, $x=5-2t, \quad y=1+(b-1)t, \quad z=a+(1-a)t.$

The line crosses the $yz$-plane when $x=0$. So, $5-2t=0 \implies t=\frac{5}{2}.$

At this value of $t$, the point of intersection is given to be $\left(0,\frac{17}{2},-\frac{-13}{2}\right).$ Note that $-\frac{-13}{2}=\frac{13}{2}.$ So the point is $\left(0,\frac{17}{2},\frac{13}{2}\right).$

Use the $y$-coordinate: $1+(b-1)\cdot \frac{5}{2} = \frac{17}{2}.$ Multiply by $2$: $2+5(b-1)=17.$ $2+5b-5=17$ $5b-3=17$ $5b=20$ $b=4.$

Use the $z$-coordinate: $a+(1-a)\cdot \frac{5}{2} = \frac{13}{2}.$ Multiply by $2$: $2a+5(1-a)=13.$ $2a+5-5a=13$ $5-3a=13$ $-3a=8$ $a=-\frac{8}{3}.$

This gives $a=-\frac{8}{3}, \quad b=4.$ This does not match any of the options.

Since the stored answer is option C: $a=6,b=4$, let us check whether the $z$-coordinate in the question may have a sign typo.

If the intersection point were $\left(0,\frac{17}{2},-\frac{13}{2}\right),$ then from $a+(1-a)\cdot \frac{5}{2}=-\frac{13}{2}$ we get $2a+5(1-a)=-13$ $2a+5-5a=-13$ $5-3a=-13$ $-3a=-18$ $a=6.$ Together with $b=4$, this gives option C.

Therefore, the printed coordinate $-\frac{-13}{2}$ simplifies to $+\frac{13}{2}$, and using it literally yields no listed option. The stored answer C is consistent only if the intended point was $\left(0,\frac{17}{2},-\frac{13}{2}\right).$