Find the equation of a line in a 3D space which crosses two another lines

Question

I have a task to find the equation of a line in a three-dimensional space which crosses two another lines with equations (on photo). We also have a point A(1, 2, 1) which is on our line.

I thought when I multiply direction vectors of lines k and l I will get another vector which will be direction vector of our finding line.

I know my solution is wrong. Why I can't use this rule here?

Correct answer is:

Update: Ok. I have an idea why it's wrong. If this two lines will be in the same plane, then the cross product won't be direction vector of our line, right?

Answer 1

HINT. Find the equation of the plane passing through $A$ and $l$, then find its intersection $B$ with line $k$. Line $AB$ is the requested line.

Answer 2

Let L' be the desired line. We mut find a vector in the direction of L'.Since L and L' intersect, there is a plane, which we'll call $\Pi _1$, containing L and L' Similarly there is a plane,which we'll call $\Pi _2$ containing K and L'.The points (1,2,1) and (1,-3,1) are both in the plane $\Pi _1$ so we subtract their coordinates to obtain the vector [0,5,0] in the plane $\Pi _1$. The vector [1,-2,2] is also in the plane $\Pi _1$. Thus the vector $$[0,5,0] \times [1,-2,2] $$ is normal to the plane $\Pi _1$The points (1,2,1) and (2,2,0) are both in the plane $\Pi _2$ so we subtract their coordinates to obtain the vector [-1,0,1] in the plane $\Pi _2$. The vector [2,1,3] is also in the plane $\Pi _2$. Thus the vector $$[-1,0,1] \times [2,1,3] $$ is normal to the plane $\Pi _2$The line L' lies in both planes, so a vector in the direction of L', which we'll call $[a,b,c]$ must be perpendicular to the normals to both planes. Thus $$[a,b,c]=([0,5,0] \times [1,-2,2]) \times ([-1,0,1] \times [2,1,3]) $$ The equation of L' is $x=1+aT,y=2+bT,z=1+cT.$