How do we express the equation corresponding to the intersection of the planes $x + y + z = 1$ and $x + 2y + 2z = 0$? If a system with three unknowns and two equations are such that
$$
\begin{align}
x+y+z=1&\\
x+2y+2z=0
\end{align}
$$
In the answer it says that this system can be represented as
$$
\begin{pmatrix}
x \\ y \\z
\end{pmatrix} = \begin{pmatrix}
2 \\-1\\0
\end{pmatrix}
+t\begin{pmatrix}
0\\-1\\1
\end{pmatrix}, t\in{\\R}
$$
But I am confused on how they came up to this parametric equation. Is there a general way of coming up with this parametric form?
 A: The proposed system of equations corresponds to the intersection of two planes which are not parallel, also known as a line. In order to obtain one of its possible parametrizations, subtract the first equation from the second in order to obtain:
\begin{align*}
(x + 2y + 2z) - (x + y + z) = 0 - 1 & \Longleftrightarrow y + z = -1
\end{align*}
Hence we conclude that $x = 2$. Consequently, if we let $z = t\in\mathbb{R}$, it results that $y = -1 - t$.
Finally, one gets the desired equation of the intersecting line:
\begin{align*}
\begin{pmatrix}
x\\
y\\
z
\end{pmatrix} =
\begin{pmatrix}
2\\
-1 - t\\
t
\end{pmatrix} =
\begin{pmatrix}
2\\
-1\\
0
\end{pmatrix} + t
\begin{pmatrix}
0\\
-1\\
1
\end{pmatrix}
\end{align*}
Hopefully this helps!
A: In order to answer the following question:

But I am confused on how they came up to this parametric equation. Is there a general way of coming up with this parametric form?

A line $L\in {\bf R}^{3}$ can always be presented as the intersection of two non-parallel planes, yielding a system of equations of the form $$L:\begin{cases}a_{1}x+b_{1}y+c_{1}z=d_{1},\\a_{2}x+b_{2}y+c_{2}z=d_{2} \end{cases}.$$
For this system of equations determine a line $L$ in ${\bf R}^{3}$, we need that ${\rm rank}\begin{pmatrix}a_{1}&b_{1}&c_{1}\\a_{2}&b_{2}&c_{2} \end{pmatrix}=2$ or in other words we need the normal vectors $\overrightarrow{n}_{1}=(a_{1},b_{1},c_{1})$ and $\overrightarrow{n}_{2}=(a_{2},b_{2},c_{2})$ not being proportional, i.e., $\frac{a_{1}}{a_{2}}\not=\frac{b_{1}}{b_{2}}$ or $\frac{a_{1}}{a_{2}}\not=\frac{c_{1}}{c_{2}}$ or $\frac{b_{1}}{b_{2}}\not=\frac{c_{1}}{c_{2}}$.
Then, that we have verified that the system of equations generates a line in three dimensional space, there are two elementary ways to find that line. The first algebraic form was give by the Átila Correia's answer and the other way a little more vectorial but still algebraic was give by Doug M's comment.
I am going to expand the second way:

*

*First notice that ${\rm rank}\begin{pmatrix}1&1&1\\1&2&2 \end{pmatrix}=2$, then the system of equations generate a line in the three dimensional space.

*In order to find a equation of a line $L$ in the three dimensional space we need: i) a point $P\in L$ and ii) a director vector $\overrightarrow{s}$ for $L$.

*The director vector $\overrightarrow{s}$ for the line $L$ can be find using the cross vector (think about it fact) between the normal vector $\overrightarrow{n}_{1}=(1,1,1)$ and $\overrightarrow{n}_{2}=(1,2,2)$. In this case we have $$\overrightarrow{s}:=\overrightarrow{n}_{1}\times \overrightarrow{n}_{2}:=\det\begin{pmatrix}\vec{i}&\vec{j}&\vec{k}\\1&1&1\\1&2&2\end{pmatrix}=(0,-1,1).$$

*The point $P$ lying on the line $L$ can be find taking a particular value for $x$ or $y$ or $z$ and solving the remaining system of equations. For example, letting $z=0$, we solve $\begin{cases}x+y=1,\\x+2y=0\end{cases}$ give $y=-1$ and then $x=2$. Then a point $P$ on the line $L$ is given by $P(2,-1,0)$ (question for you: what if we choose $y=0$, for example?)

*Every equation of a line in the three dimensional space can be write as $L(t)=P+t\overrightarrow{s}$ for $t$ a real parameter. Thus, the line is given by

$$L(t)=(2,-1,0)+t(0,-1,1),\quad t\in {\bf R}$$
or logically equivalently in  order of your notation as
$$\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}2\\-1\\0\end{pmatrix}+t\begin{pmatrix}0\\-1\\1\end{pmatrix}$$
for $(x,y,z)\in {\bf R}^{3}$ and $t\in {\bf R}$.
