Q.Find the Cartesian equation of the line passing through the origin which is perpendicular to $x - 1 = y - 2 = z - 1$ and intersects the $\dfrac{x-1}{2} = \dfrac{y+1}{3} = \dfrac{z-1}{4}$.

[!TLDR] The required line's direction $(a,b,c)$ satisfies $a+b+c=0$ (perpendicular to $x-1=y-2=z-1$) and is proportional to the intersection point with the second line. …

Let the required line, through the origin with direction ratios $(a,b,c)$, meet $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-1}{4}=t$ at the point $Q=(1+2t,\ -1+3t,\ 1+4t)$.

Since the required line passes through the origin and through $Q$, its direction is proportional to $Q$ itself, i.e. $(a,b,c)\propto(1+2t,\ -1+3t,\ 1+4t)$.

Perpendicularity with $x-1=y-2=z-1$ (direction $(1,1,1)$) requires $a+b+c=0$; since $(a,b,c)$ is proportional to $Q$, this same condition applies to $Q$'s coordinates:

So …

Parametrise a point $Q$ on the second line, use the fact that the required line through the origin and $Q$ has direction proportional to $Q$, impose the perpendicularity condition $a+b+c=0$ on …

Forgetting that a line through the origin has direction exactly proportional to any other point on it, so the perpendicularity condition can be applied directly to $Q$' …