The point of intersection of the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ is:

First line Let parameter $t$ $x=1+2t$ $y=2+3t$ $z=3+4t$ Second line Let parameter $s$ $\frac{x-4}{5}=\frac{y-1}{2}=z=s$ So $x=4+5s$ $y=1+2s$ $z=s$ For intersection $3+4t=s$ Substitute in $x$ $1+2t=4+5(3+4t)$ $1+2t=19+20t$ $-18=18t$ $t=-1$ Then $s=3+4(-1)=-1$ Point $x=1+2(-1)=-1$ $y=2+3(-1)=-1$ $z=3+4(-1)=-1$