Plane determined by two lines 
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"?
 A: If two lines in 3D space ($\Bbb R^3$) intersect or are parallel there is a plane in that 3D space that contains those two lines.
So you can define a plane by defining two lines that intersect or are parallel.
A: Rewrite the lines in vector form:
$$l_1=(-2,3,4)+\lambda(1,2,-1)$$
$$l_2=(3,4,0)+\mu(-1,-2,1)$$
where $\lambda,\mu\in\Bbb R$.
Since $(1,2,-1)=-1\cdot(-1,-2,1)$, either the two lines are coincident or they are parallel. Since $(3,4,0)$ does not lie on $l_1$, they are parallel.
To find the plane containing both lines, take the vector connecting their initial points: $(3,4,0)-(-2,3,4)=(5,1,-4)$. Now do the cross product of this vector with either of the line's direction vectors: $(5,1,-4)×(1,2,-1)=(7,1,9)$. This gives the plane's normal, and all that remains is to compute the scalar of the plane equation: $(7,1,9)\cdot(3,4,0)=25$.
Therefore the plane containing both lines has the equation
$$\mathbf r\cdot(7,1,9)=25$$
or in Cartesian form
$$7x+y+9z=25.$$
A: For $t=0$ and $t=1$ we get two points in each line.
In $L_1$ we get $(-2,3,4)$ and $(-1,5,3)$.
In $L_2$ we get $(3,4,0)$ and $(2,2,1)$.
It follows the a same director vector, $\vec u=\vec i+2\vec j-\vec k$, for both lines which means the lines are parallel and obviously distinct so its determine the plan containing both lines.

A: the parametric equations of two lines are
For first line
$$\frac{x+2}{1}=\frac{y-3}{2}=\frac{z-4}{-1}$$ 
For second line
$$\frac{x-3}{-1}=\frac{y-4}{-2}=\frac{z-0}{1}$$ 
multiply the second parametric equation by $(-1)$
$$\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-0}{-1}$$ 
the Denominators for both equations are equal, so the two lines are parallel.
the general equation of plane is
$$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$
so that the $A$,$B$, and $C$ are the components of normal vector on plane
to find the normal vector:
$$\overrightarrow{v_1}=i+2j-k$$
$$\overrightarrow{v_2}=(-2-(3))i+(3-(4))j+(4-0)k$$
$$\overrightarrow{v_2}=-5i-j+4k$$
the normal vector is
$$\overrightarrow{n}=\overrightarrow{v_1}\times \overrightarrow{v_2}$$
then comlplete the solution
A: 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.    
$7x+y+9z=25$
A: take any one point from two lines and find the equation of line for this and find the perpendicular vector by cross product of the give any line and the new line this will be enough to find the plane
