$x=p y+q, z=r y+s$here, we put y = t$x=p t+q, z=r t+s$$(x, y, z)=(p t+q, t, r t+s)$Direction vector: $\vec{d}_1 = (p, 1, r)$Line 2:$x=p' y+q', z=r' y+s'$Here, we put $y=t'$$x=p' t'+q', z=r' t'+s'$$(x, y, z)= (p' t'+q', t', r' t'+s')$Direction vector: $\vec{d}_2 = (p', 1, r')$Two lines are perpendicular when dot product of their direction vectors is zero.$\vec{d_1} \cdot \vec{d_2} = 0$$p p'+1+r r'=0$$p p'+r r'=-1$, but in assertion it is given that $p p'+r r'=1$. So assertion is false.Two lines $\vec{r} = \vec{a_1} + \lambda \vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu \vec{b_2}$ are perpendicular when the dot product of their direction vector is zero.Here, direction vectors are $\vec{b_1}$ and $\vec{b_2}$ respectively.So $\vec{b_1} \cdot \vec{b_2} = 0$Reason is true.