equation of a plane passing through a line and parallel to another line verification In homogeneous coordinates I have the points $M(2,3,1,1), N(3,3,2,1), P(1,3,1,1), Q(0,2,1,0)$.
I have to find equation of a plane which contains the line PQ and it's parallel to MN.
What I have so far:
$$PQ: 
\left\{ 
\begin{array}{c}
x=1\mu+0\lambda \\ 
y=3\mu+2\lambda \\ 
z=1\mu+1\lambda \\ 
t=1\mu+0\lambda
\end{array}
\right. 
\text{ and } MN: 
\left\{ 
\begin{array}{c}
x=2\mu+3\lambda \\ 
y=3\mu+3\lambda \\ 
z=1\mu+2\lambda \\ 
t=1\mu+1\lambda
\end{array}
\right. 
$$
From the first answer of this question I know that I can form a determinant with rows the directions of the lines.
$$
\begin{array}{|aaa|}
x & y & z\\
0& 2 & 1\\
3 & 3 & 2\\
\end{array}=0
$$
From where
$\alpha: x-3y+3z-a=0$.
Since $\alpha$ passes through $P(1,3,1,1)$ I substitute it's coordinates in the equation of $\alpha$ and I get:
$\alpha: x-3y+3z+5=0$.
Is my reasoning correct? I have the feeling that being in homogeneous coordinates the approach is entirely different.
Thanks in advance!
 A: The plane we are looking for in the projective space corresponds to a 3d linear subspace of $\Bbb R^4$. 
It contains $P$ and $Q$, and it contains the ideal point $I$ of the line $MN$, which is now $I=N-M=(1,0,-1,0)$. 
We thus have $3$ vectors of $\Bbb R^4$, and are looking for a common orthogonal vector to them (the coefficients of the equation of the plane). 
For this, the same determinant formula can be applied:
$$\left\vert\matrix{x&y&z&1\\ 1&3&1&1\\ 0&2&1&0\\ 1&0&-1&0}\right\vert\ =\ 0$$
A: With $p = (x,y,z,t)^{\top}$ given the lines
$$
L_{PQ}\to p = P + \lambda_1(P-Q)\\
L_{MN}\to p = M + \lambda_2(M-N)
$$
and the plane 
$$
\Pi\to < p-p_0, \vec n > = 0
$$
If $L_{PQ}\in \Pi\Rightarrow < P-p_0+\lambda_1(P-Q),\vec n > = 0$ or $< P-p_0,\vec n > = 0$ and $ < P-Q, \vec n > = 0$
If $L_{MN} || \Pi \Rightarrow < M-N, \vec n > = 0$ then we have
$$
\vec n \perp (P-Q)\\
\vec n \perp (M-N)\\
<P,\vec n > = < p_0,\vec n >
$$
then choosing $p_0 = P$ we have the plane equation as
$$
\Pi\to < p-P,\vec\omega> = 0
$$
where $\vec\omega$ can be one of the solutions for
$$
< P-Q,\vec \omega > = 0\\
< M-N, \vec \omega > = 0\\
||\vec\omega|| = 1
$$
for instance
$$
\vec\omega = \left\{\frac{1}{3} \left(-\sqrt{3-5 \lambda ^2}-\lambda \right),\frac{1}{3} \left(\sqrt{3-5 \lambda ^2}-2 \lambda \right),\frac{1}{3}
   \left(\sqrt{3-5 \lambda ^2}+\lambda \right),\lambda \right\}
$$
for any $\lambda \in \Re$ such that $3-5\lambda^2 \ge 0$
