The two lines $x=a y+b, z=c y+d ;$ and $x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime}$ are…
The two lines $x=a y+b, z=c y+d ;$ and $x=a^{\prime} y+b^{\prime}, z=c^{\prime} y+d^{\prime}$ are perpendicular to each other if
$\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=-1$
$\mathrm{aa}^{\prime}+\mathrm{cc}^{\prime}=1$
$\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=-1$
$\frac{a}{a^{\prime}}+\frac{c}{c^{\prime}}=1$
Solution
Equation of lines $\frac{x-b}{a}=y=\frac{z-d}{c}$
$\frac{x-b^{\prime}}{a^{\prime}}=y=\frac{z-d^{\prime}}{c^{\prime}}$
Lines are perpendicular $\Rightarrow a a^{\prime}+1+c c^{\prime}=0$