The symmetrical form of the line of intersection of the planes $ x = ay +b, z=cy + d, $ is

$\frac{x-a}{a}=\frac{y-0}{1}=\frac{z-d}{c}$

Let the direction ratios of the required line be proportional to $l, m, n.$ Since the required line lies in both the given planes. Therefore,

$l + m(-a) + n × 0= 0$

and, $l × 0 + m(-c) + n = 0 $

Solving these two equations by cross-multiplication, we get

$\frac{l}{-a}=\frac{m}{-1}=\frac{n}{-c}⇒ \frac{l}{a}=\frac{m}{1}=\frac{n}{c}$

In order to obtain the coordinates of a point on the required line, we put y = 0 in the two given equations to obtain 

$x= b, z = d $.

Thus, the coordinates of a point on the required line are (b, 0, d).

Hence, its required is $\frac{x-a}{a}=\frac{y-0}{1}=\frac{z-d}{c}$