Intersection of 3D lines

Question

How to find intersection point of lines $p... \frac{x + 1}{0} = \frac{y - 2}{-1} = \frac{z}{1}$ and $q... \frac{x - 1}{-1} = \frac{y + 6}{3} = \frac{z + 6}{4}$?

Answer 1

Let $\frac{x + 1}{0} = \frac{y - 2}{-1} = \frac{z}{1}=t\in\mathbb{R}$. Then $x=-1, y=2-t, z=t$. So, the parametric form of the first line is $c_1(t)=(-1,2-t,t),\ t\in\mathbb{R}$. If we do the same for the second line, we get it's parametric form $c_2(t)=(1-t, 3t-6,4t-6),\ t\in\mathbb{R}$. Thus, $$c_1(t)=c_2(t) \iff t=2. $$ Consequently, the intersection point is $A(-1,0,2)$.