The lines $x=py+q,z=ry+s$ and $x=p^{\prime }y+q^{\prime },z=r^{\prime }y+s^{\prime }$ are perpendicular to each other, if:

Rearranging the equations of the lines in symmetrical form:For the first line: $x-q=py⟹\frac{x-q}{p}=\frac{y}{1}$ and $z-s=ry⟹\frac{z-s}{r}=\frac{y}{1}$.Line 1: $\frac{x-q}{p}=\frac{y}{1}=\frac{z-s}{r}$. Direction ratios $(a_1,b_1,c_1)=(p,1,r)$.For the second line: $x-q^{\prime }=p^{\prime }y⟹\frac{x-q^{\prime }}{p^{\prime }}=\frac{y}{1}$ and $z-s^{\prime }=r^{\prime }y⟹\frac{z-s^{\prime }}{r^{\prime }}=\frac{y}{1}$.Line 2: $\frac{x-q^{\prime }}{p^{\prime }}=\frac{y}{1}=\frac{z-s^{\prime }}{r^{\prime }}$. Direction ratios $(a_2,b_2,c_2)=(p^{\prime },1,r^{\prime })$.For the lines to be perpendicular, the dot product of their direction vectors must be zero:$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$$(p)(p^{\prime })+(1)(1)+(r)(r^{\prime })=0$$p{p}^{\prime }+r{r}^{\prime }+1=0$