The line passing through the points $\mathrm{A}(5,1,\mathrm{a})$ and $B(3,b,1)$ crosses $yz-$Plane at the point  $\left(\text{0,}\frac{\text{17}}{\text{2}}\text{,}\frac{\text{-13}}{\text{2}}\right)$  then the values of a,b are 

yz – plane divides $\overline{\text{AB}}$  in the ratio  = -x1;x2= -5:3

$$\begin{gathered} \therefore \left(\text{0,}\frac{\text{-5b+3}}{\text{-2}}\text{,}\frac{\text{-5+3a}}{\text{-2}}\right)\text{=}\left(\text{0,}\frac{\text{17}}{\text{2}}\text{,}\frac{\text{-13}}{\text{2}}\right) \\ \Rightarrow \frac{-5b+3}{-2}=\frac{17}{2};\frac{-5+3a}{-2}=\frac{-13}{2} \\ \Rightarrow 5b-3=17;\Rightarrow -5+3a=13 \\ \Rightarrow 5b=20;3a=18 \\ \Rightarrow b=4;\Rightarrow a=6 \\ \therefore \text{a = 6,b = 4} \end{gathered}$$