Step 1: Parametrise both lines.
Let $y=t$ on the first line. A general point is $(pt+q,\ t,\ rt+s)$. For each unit change of $t$, the point moves by the vector $\vec{d_1} = p\,\hat{i} + \hat{j} + r\,\hat{k}$.
Step 2: Do the same for the second line.
Let $y=u$. A general point is $(p'u+q',\ u,\ r'u+s')$. The direction vector is $\vec{d_2} = p'\hat{i} + \hat{j} + r'\hat{k}$.
Step 3: Use the angle formula.
The angle $\theta$ between the lines satisfies $\cos\theta = \dfrac{\vec{d_1}\cdot\vec{d_2}}{|\vec{d_1}||\vec{d_2}|}$. Perpendicular lines have $\theta=90^\circ$, so the numerator must vanish.
Step 4: Compute the dot product.
\[ \vec{d_1}\cdot\vec{d_2} = pp' + 1 + rr' \]
Setting this to zero gives $pp'+rr'+1=0$.
Step 5: Match with the options.
Only option 2 has the extra $+1$. Options 1 and 4 miss the y-components. Option 3 is a proportionality relation, not a zero dot product.
Final Answer:
The condition is $pp'+rr'+1=0$.
\[ \boxed{\text{Option 2}} \]