1. Write each line in parametric/vector form

The first line is given by $x=ay+b,\qquad z=cy+d.$ Let us take $y=t$ as parameter. Then $x=at+b,\quad y=t,\quad z=ct+d.$ So its direction ratios are $\vec{d_1}=(a,1,c).$

The second line is given by $x=a'z+b',\qquad y=c'z+d'.$ Let us take $z=s$ as parameter. Then $x=a's+b',\quad y=c's+d',\quad z=s.$ So its direction ratios are $\vec{d_2}=(a',c',1).$

2. Condition for perpendicular lines

Two lines are perpendicular if their direction vectors are perpendicular, i.e. their dot product is zero: $\vec{d_1}\cdot\vec{d_2}=0.$

So, $(a,1,c)\cdot(a',c',1)=0.$

Compute the dot product: $aa' + 1\cdot c' + c\cdot 1 = 0.$ Hence, $aa' + c' + c = 0.$

That is, $aa' + c + c' = 0.$

3. Match with the options

This corresponds exactly to:

D: $aa' + c + c' = 0$

4. Check other options

The constants $b,b',d,d'$ only affect the position of the lines, not their directions. Therefore any condition for perpendicularity cannot involve $b$ or $b'$. So options A and C are immediately invalid.

Also option B gives $cc' + a + a' = 0,$ which does not come from the dot product of the direction vectors.

Therefore the correct option is D.

5. Comparison with stored answer

Stored correct answer: D

My derived answer: D

They agree.