Let's take the two points A(3,2,-1) and B(1,-1,2).
Let's use the point A(3,2,-1) as the fixed point on the line. We need a direction vector that is parallel to the vector AB.
\(\displaystyle V=AB=(1-3)i+(-1-2)j+(2-(-1))k=-2i-3j+3k\)
The symmetric equations are \(\displaystyle \frac{x-3}{-2}=\frac{y-2}{-3}=\frac{z+1}{3}\)
The parametric equations are \(\displaystyle x=3-2t, \;\ y=2-3t, \;\ -1+3t\)
The plane contains the line:
\(\displaystyle x=3-2t, \;\ y=2-3t, \;\ z=-1+3t\)
The plane contains the point (3,2,-1) and the normal of the plane must be orthogonal to the direction vector
\(\displaystyle V_{1}=-2i-3j+3k\)
Since the plane is parallel to the line \(\displaystyle x=1+3t, \;\ y=-1+2t, \;\ z=-t\) the normal to the plane must also be orthogonal to a direction vector of that line, \(\displaystyle V_{2}=3i+2j-k\)
The cross product \(\displaystyle V_{1}\times V_{2}=-3i+7j+5k\) is the normal to the plane.
Find the equation of the plane that contains the point (3,2,-1) with normal vector -3i+7j+5k
So, the plane has equation \(\displaystyle -3(x-3)+7(y-2)+5(z+1)=0\)
\(\displaystyle -3x+7y+5z=0\)
Let's now check to see if the lines don't intersect.
We have parametric equations of the lines as:
\(\displaystyle x=3-2t, \;\ y=2-3t, \;\ z=-1+3t\)
They must be expressed in different parameters, so the other line can be expressed as:
\(\displaystyle x=1+3s, \;\ y=-1+2s, \;\ z=-s\)
Set corresponding coordinates equal on each line and we get the system:
\(\displaystyle \left\{ \begin{array} \text{3-2t}=1+3s\\ 2-3t=-1+2s\\ -1+3t=-s\end{array}\) (I do not know why that little 'c' is at the top. Disregard).
The first two equations have solution \(\displaystyle (t,s)=(1,0)\)
Since these values do not satisfy the third equation, the lines are parallel.
Check my 'figgers' to make sure I did not go down the primrose path. Easy to do in all that.