Equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$

Question

Find the equations of the line which intersects the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z+1}{4}$ and passes through the point $(1,1,1)$.

First I thought required line passes through point on intersection of given lines, but as I checked, lines are non-intersecting. So doesn't that mean that there will be infinite lines which satisfy the given conditions?

Answer 1

Let us choose $P_1(2t+1,3t+2,4t+3)$ on the first line, and $P_2(s-2,2s+3,4s-1)$ on the second line.

the point $P(1,1,1)$ is known.

So let $P_1,P_2,P$ on the same line, i.e.,

solving $|\vec{P P_1} \times \vec{P P_2}| = 0$ to get the relation between $s$ and $t$ (maybe even the exact values of them)

Answer 2

write the first equation in the form $$ [x,y,z]=[1,2,3]+t[2,3,4]$$ and the second one in the form $$[x,y,z]=[-2,3,-1]+s[1,2,4]$$

Answer 3

Can't comment:

To get to the result that J. Wu and Dr. Sonnard Graubner gets, you can assume that everything is equal to $t$ or seperately

$$t=\frac{x-1}{2}; t=\frac{y-2}{3};t=\frac{z-3}{4}$$

and solve for $x,y,$ and $z$ in each seperate equations