Hi, another vector question I'm afraid

I have the 4 lines:

$\displaystyle BC = \begin{pmatrix} -2 \\ -4 \\ 1 \end{pmatrix} + \lambda\begin{pmatrix} 3 \\ 6 \\ -3 \end{pmatrix}$, $\displaystyle AC = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} + \sigma\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}$.

$\displaystyle l2 = \begin{pmatrix} 3 \\ 0 \\ 2 \end{pmatrix} + \beta\begin{pmatrix} -2 \\ 2 \\ -4 \end{pmatrix}$, $\displaystyle AD = \begin{pmatrix} 0 \\ 0 \\ -1 \end{pmatrix} + \alpha\begin{pmatrix} -3 \\ 0 \\ -3 \end{pmatrix}$.

I need to show that all 4 lines lie in a single plane and fine the Cartesian Equation of the plane.

The Cartesian equation of a plane is in the form $\displaystyle Ax+By+Cz=d$, so from this I can set up the following equations:

$\displaystyle -2A-4B+C=d$

$\displaystyle 3A+2C=d$

$\displaystyle A+2B-2C=d$

From these I can get all $\displaystyle A,B,C$ in terms of $\displaystyle d$.

$\displaystyle A=d$, $\displaystyle B=C=-d$

Putting these values back into the original Cartesian Equation I got:

$\displaystyle dx-dy-dz = d$, which can be simplified to $\displaystyle x-y-z=1$.

Is this correct so far or have I gone horribly wrong?

Thanks