Find the equation of the plane that passes through two lines. Find the equation of the plane that passes through

                                               x = 5 + 3t
                                               y = 1
                                               z = -t

and

                                              x = 4 + 3t
                                              y = 2 + 3t
                                              z = 1 - t



Can someone explain why these lines dont intersect, and therefore dont produce a plane equation...
the point of intersection of both lines is when x=5+3t = 4+3t 
 A: Well, looking at the first equation, y = 1 (constant).  So if the lines intersect it will be at y = 1.  So solve the y in the second equation for 1.
$y = 2 + 3t \implies 1 = 2 + 3t \implies -1 = 3t \implies \frac{-1}{3} = t$
If we plug in that t value for the z equations, we see they are unequal.
The first equation:
$z = -t \implies z = \frac{1}{3}$
The second equation:
$z = 1 - t \implies z = 1\frac{1}{3} $
Since, since the lines can never share the same y value and the same z value at the same time, they do not intersect.
That said, since you tagged this problem [linear algebra] and [vector spaces] the work should probably be done with matrices.
A little post script, after seeing a comment that there is no such plane that intersects both lines.  Sorry to answer here I lack the points to comment.  That answer is ether poorly worded or wrong.  There are an infinite number of planes that pass through both lines. Take any point from each line and any other point not colinear with the first two, and you have defined a plane that intersects with the lines at the points you chose.
The answer should be phrased, there exists no plane in which both lines lie.
