If two lines $$L_1$$ and $$L_2$$ in space, are defined by
$$L_1 = \{x = \sqrt{\lambda}y + (\sqrt{\lambda} - 1), z = (\sqrt{\lambda} - 1)y + \sqrt{\lambda}\}$$ and
$$L_2 = \{x = \sqrt{\mu}y + (1 - \sqrt{\mu}), z = (1 - \sqrt{\mu})y + \sqrt{\mu}\}$$
then $$L_1$$ is perpendicular to $$L_2$$, for all nonnegative reals $$\lambda$$ and $$\mu$$, such that :

Using $$y$$ as the parameter for both lines, the direction ratios of $$L_1$$ are obtained from the coefficients of $$y$$ in each coordinate equation. For $$L_1$$: $$x = \sqrt{\lambda}\,y + (\sqrt{\lambda}-1)$$, $$y = y$$, $$z = (\sqrt{\lambda}-1)\,y + \sqrt{\lambda}$$, so the direction ratios are $$(\sqrt{\lambda},\; 1,\; \sqrt{\lambda}-1)$$. Similarly for $$L_2$$: $$x = \sqrt{\mu}\,y + (1-\sqrt{\mu})$$, $$y = y$$, $$z = (1-\sqrt{\mu})\,y + \sqrt{\mu}$$, giving direction ratios $$(\sqrt{\mu},\; 1,\; 1-\sqrt{\mu})$$.

For perpendicularity the dot product of the direction vectors must vanish: $$\sqrt{\lambda}\cdot\sqrt{\mu} + 1\cdot 1 + (\sqrt{\lambda}-1)(1-\sqrt{\mu}) = 0$$. Expanding the third term: $$\sqrt{\lambda} - \sqrt{\lambda}\sqrt{\mu} - 1 + \sqrt{\mu}$$. Adding all three parts gives $$\sqrt{\lambda\mu} + 1 + \sqrt{\lambda} - \sqrt{\lambda\mu} - 1 + \sqrt{\mu} = \sqrt{\lambda} + \sqrt{\mu} = 0$$.

Since $$\lambda$$ and $$\mu$$ are nonnegative reals, $$\sqrt{\lambda} \geq 0$$ and $$\sqrt{\mu} \geq 0$$, so their sum equals zero only when both are zero, i.e. $$\lambda = \mu = 0$$. This is equivalent to the condition $$\lambda = \mu$$ (both equal to zero).