The lines x = ay + b, z = cy + d and x = a’y + b’, z = c’y + d’ will be mutually perpendicular provided :

(a + a’)(b + b’)(c + c’) aa’ + cc’ + 1 = 0 aa’ + bb’ + cc’ + 1 = 0 (a + a’) (b + b’) (c + c’) + 1 = 0

aa’ + cc’ + 1 = 0

Give lines are $\frac{x-b}{a}=y=\frac{z-d}{c}$  and  $\frac{x-b'}{a'}=y=\frac{z-d'}{c'}$ These lines will be mutually perpendicular, provided a.a' + 1.1' + c.c' =0 ⇒ a.a' + c.c' + 1 = 0