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?

Maybe you did not report the whole question.
This solution gives the equation of some plane that contains the given line. Indeed,
$$a=b=c\implies a+b=2c,$$ but there are infinitely many other combinations, like
$$a=b=c\implies 7a-6b=c$$ or $$a=b=c\implies a=b,$$
all giving planes.
