The two lines $$x = ay + b, z = cy + d$$; and $$x = a'y + b', z = c'y + d'$$ are perpendicular to each other if

For each line, rewrite the given pair of equations in the form of a single vector equation so that the direction ratios (d.r.’s) are obvious.

Line 1: $$x = ay + b,\; z = cy + d$$ can be written as
$$(x,\;y,\;z) = (b,\;0,\;d) + y\,(a,\;1,\;c).$$
Hence the direction vector of this line is $$\mathbf{v}_1 = (a,\;1,\;c).$$

Line 2: $$x = a'y + b',\; z = c'y + d'$$ becomes
$$(x,\;y,\;z) = (b',\;0,\;d') + y\,(a',\;1,\;c').$$
So its direction vector is $$\mathbf{v}_2 = (a',\;1,\;c').$$

Two lines are perpendicular when their direction vectors are perpendicular; that is, the dot product $$\mathbf{v}_1 \cdot \mathbf{v}_2$$ equals zero.

Compute the dot product:
$$\mathbf{v}_1 \cdot \mathbf{v}_2 \;=\; (a,\;1,\;c) \cdot (a',\;1,\;c') \;=\; aa' + 1\cdot1 + cc'.$$

For perpendicularity we require
$$aa' + 1 + cc' = 0 \;\Longrightarrow\; aa' + cc' = -1.$$

Therefore the correct condition is $$aa' + cc' = -1,$$ which corresponds to Option A.

Final answer: Option A which is: $$aa' + cc' = -1$$