TITLE: Plane determined by two lines
QUESTION [5 upvotes]: Show that the lines $x=-2+t,y=3+2t,z=4-t$ and $x=3-t,y=4-2t,z=t$ are parallel. Find the equation of the plane they determine.

Here what is the meaning of "they determine"?

REPLY [1 votes]: 1947898
Coëfficients of $t\text{ in first line:}\{1,2,-1\}$
Coëfficients of $t\text{ in second line:}\{-1,-2,1\}$
These are the top two rows of a determinant of order 3.
The third row is: $\{1,1,1\}.$
The determinant    
$\left|\begin{array}{rrr}
1&2&-1\\
-1&-2&1\\
1&1&1
\end{array}\right|=0,$
so the two lines are parallel.
Compute two points from the first line, say,
$(-2\mid 3\mid 4)\text{ and }(-3\mid 1\mid 5)$
and one point from the second, say, $(3\mid 4\mid 0).$
The equation    
$\left|\begin{array}{rrrr}
x&y&z&1\\
-2&3&4&1\\
-3&1&5&1\\
3&4&0&1
\end{array}\right|=7x+y+9z-25=0$
is the equation you seek.    
$7x+y+9z=25$
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