Intersections points of two straights in its symmetrical shape

Question

Find the intersection points of the straights

$$L_1:\,\,\,\,\,\frac{2-x}{3}=y+4=z$$ $$L_2:\,\,\,\,\,x+1=\frac{-y-3}{4}=\frac{1-z}{6}$$

The exercise should I solve without using the parametric representation of the straight

Answer 1

From the first line, we have $z=y+4$; and from the second line, we have $-3y-9=2-2z$

so $2z=11+3y$. Then $2(y+4)=11+3y\implies 2y+8=11+3y\implies y=-3$, so

$z=(-3)+4=1$ and $x+1=\frac{3-3}{4}=0\implies x=-1$.

Therefore $(-1,-3,1)$ is the point of intersection of the two lines.