Equation of line through $$\left( {5,1,a} \right)$$ and

$$\left( {3,b,1} \right)$$ is $${{x - 5} \over { - 2}} = {{y - 1} \over {b - 1}} = {{z - a} \over {1 - a}} = \lambda $$

$$\therefore$$ Any point on this line is a

$$\left[ { - 2\lambda + 5,\left( {b - 1} \right)\lambda + 1,\left( {1 - \alpha } \right)\lambda + a} \right]$$

It crosses $$yz$$ plane where $${ - 2\lambda + 5 = 0}$$

$$\lambda = {5 \over 2}$$

$$\therefore$$ $$\left( {0,\left( {b - 1} \right){5 \over 2} + 1,\left( {1 - a} \right){5 \over 2} + a} \right) = \left( {0,{{17} \over 2},{{ - 17} \over 2}} \right)$$

$$ \Rightarrow \left( {b - 1} \right){5 \over 2} + 1 = {{17} \over 2}$$

and $$\left( {1 - a} \right){5 \over 2} + a = - {{13} \over 2}$$

$$ \Rightarrow b = 4$$ and $$a = 6$$