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!

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:
$$\left\vert\matrix{x&y&z&1\\ 1&3&1&1\\ 0&2&1&0\\ 1&0&-1&0}\right\vert\ =\ 0$$
