How to find the cartesian equation of a plane given the vector equation? The vector equation is given by $r=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$
To find the cartesian equation, I have to consider a point  on the plane:
$$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$$
which gives:
$$\frac{x+3}{\:6}=\:\frac{y+5}{6}=\frac{z+1}{-3}$$
Now this is the part where I don't understand. Up till now my working follows with the answers provided by my book, but then the book did something that puzzle me:
$$\frac{x+3}{6}+\frac{y+5}{6}=2(\frac{z+1}{-3})$$
How did they reached here?

We have that

*

*$x=-3+6k \implies k=\frac{x+3}6$

*$y=-5+6k\implies k=\frac{y+5}6$

*$z=-1-3k\implies k=-\frac{z+1}3$
therefore
$$\frac{x+3}6+\frac{y+5}6=2k=-2\frac{z+1}3$$
moreover
$$\frac{x+3}6-\frac{y+5}6=2k=0$$
and the given line can be represented, in cartesian form, by the intersection of the two planes

*

*$x+y+4z+12=0$

*$x-y-2=0$
