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

The two given points are $$P(5,1,a)$$ and $$Q(3,b,1)$$. A line through them can be written in parametric form.

Direction ratios of $$\overrightarrow{PQ}$$:
$$\bigl(3-5,\; b-1,\; 1-a\bigr)=(-2,\; b-1,\; 1-a)$$.

Taking $$t$$ as parameter with $$t=0$$ at $$P$$, any point $$R(x,y,z)$$ on the line is
$$x = 5-2t,\qquad y = 1+(b-1)t,\qquad z = a+(1-a)t$$.

The line meets the $$yz$$-plane when $$x=0$$.
Setting $$x=0$$ gives $$5-2t=0 \Rightarrow t=\dfrac{5}{2}$$.

Substitute $$t=\dfrac{5}{2}$$ to get the coordinates of the intersection point:
$$y = 1 + (b-1)\dfrac{5}{2},\qquad z = a + (1-a)\dfrac{5}{2}.$$

This point is given as $$\left(0,\dfrac{17}{2},-\dfrac{13}{2}\right)$$, so we equate coordinates.

For the $$y$$-coordinate
$$1 + (b-1)\dfrac{5}{2} = \dfrac{17}{2}$$
$$\Rightarrow (b-1)\dfrac{5}{2} = \dfrac{17}{2}-1 = \dfrac{15}{2}$$
$$\Rightarrow 5(b-1) = 15 \Rightarrow b-1 = 3 \Rightarrow b = 4.$$

For the $$z$$-coordinate
$$a + (1-a)\dfrac{5}{2} = -\dfrac{13}{2}$$
$$\Rightarrow a + \dfrac{5}{2} - \dfrac{5a}{2} = -\dfrac{13}{2}$$
$$\Rightarrow \dfrac{2a}{2} - \dfrac{5a}{2} + \dfrac{5}{2} = -\dfrac{13}{2}$$
$$\Rightarrow -\dfrac{3a}{2} + \dfrac{5}{2} = -\dfrac{13}{2}$$
$$\Rightarrow -3a + 5 = -13 \Rightarrow -3a = -18 \Rightarrow a = 6.$$

Thus $$a = 6$$ and $$b = 4$$.

Option C which is: $$a = 6,\ b = 4$$