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?

You have $$\frac{x+3}{6}=\frac{z+1}{-3}$$ and $$\frac{y+5}{6}=\frac{z+1}{-3}$$
Now add these two equations.
