3-D Geometry - Intersection of Lines

Question

Question:

Find the equations of the line that intersects the lines

2x + y - 4 = 0 = y + 2x ; x + 3z = 4, 2x + 5z = 8

and passes through the point (2, -1, 1)

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What I tried so far

As I understand, here, the set of lines 2x + y - 4 = 0 = y + 2x and x + 3z = 4, 2x + 5z = 8 are considered as the 2 separate set of lines, which we need to show that are being intersected.

So, the equation of a general line passing through 2x + y - 4 = 0 = y + 2x is given by 2x + y - 4 + k(y + 2x) =0 for some real number k

Also, equation of a general line passing through can be represented by, x + 3z - 4 + r(2x + 5z - 8) for some real number r.

I'm naming these equations as follows for ease of use.

2x + y - 4 + k(y + 2x) = 0 ----(1)

x + 3z - 4 + r(2x + 5z - 8) = 0 -------(2)

If these lines intersect at any point(s), for some value of k and r, (1) - (2) = 0 should exist.

That is, 2x + y - 4 + k(y + 2x) - [ x + 3z - 4 + r(2x + 5z - 8) ] = 0

====> (1 - 2r)x + y(1 + k)y + ( 2k - 3 - 5r)z - 8r = 0

If I'm to solve this, I can equate coefficients of x, y, z and constants each to 0.

Which gives, (1 - 2r) = 0 and -8r = 0

How can I proceed further with this?

Answer 1

See my comment.

since two lines are in xy plane,, to intersect both with a line going through

a point, that point must have z=0.

It is nice the other two lines intersect at (4,0,0) so if the point specified had z=0

the problem is solvable. As given, the problem has no solution.