$x-y-2z=0$ find a perpendicular vector 
Why is the vector $e=(1,-1,-2)$ ?
 A: A vector $(x,y,z)$ is in the plane $P$ if we have $x-y-2z=0$, or in other words, if
$$
(1,-1,-2)\cdot(x,y,z) =0
$$
So $P$ consists of exactly those vectors that are orthogonal to $(1,-1,-2)$, just from a rewriting of the equation.
A: If $n = \langle a,b,c\rangle$ is the normal vector to the plane $P(x,y,z)$. Take the vectors $r = \langle x,y,z\rangle$ and $r_0 = \langle x_0,y_0,z_0\rangle$ such that $r - r_0 \in P(x,y,z)$ . The normal vector is orthogonal to every vector in $P(x,y,z)$. So in particular, 
$$n\dot\ (r-r_0) = 0$$
$$\langle a,b,c\rangle \langle (x-x_0),(y-y_0),(z-z_0)\rangle = 0$$ or 
$$a(x-x_0)+b(y-y_0)+c(z-z_0)= 0$$
In your case we have $$1(x-0) -1(y-0) - 2(z-0) = \langle 1,-1,-2\rangle\langle x,y,z\rangle = 0$$
A: The normal vector can be read off directly from the Cartesian equation of a plane, once the equation has been understood, the derivation of which is given by the following theorem and the accompanying corollary.



Theorem : Vector equation of a plane.
Let $\Pi$ be a plane in $\mathbb{R}^{3}$.
Let $P_{0}$ be an arbitrary point on $\Pi$.
Let $\mathbf{n}$ be a vector orthogonal to $\Pi$.
Then the vector equation of $\Pi$ is expressible as
$$\left(\mathbf{r} - \mathbf{r}_{0}\right) \cdot \mathbf{n} = 0. \qquad \qquad \qquad (1)$$
in which $\mathbf{r}$ is any point on $\Pi$ and $\mathbf{r}_{0}$ is the position vector of $P_{0}$.

Proof:
A plane in space is uniquely defined by three non-collinear points or equivalently by a point and a vector perpendicular to the plane.
Suppose $\Pi$ is a plane in space containing the points $P_{0}(x_{0},y_{0},z_{0})$, $P_{1}(x_{1}, y_{1}, z_{1})$ and $P(x, y, z)$, as illustrated in the figure. Then $\Pi$ is the set of all points $P(x, y, z)$ for which $\overrightarrow{P_{0}P_{1}}$ and $\overrightarrow{P_{0}P}$ are perpendicular to the normal vector $\mathbf{n}$. That is, from the definition of cross product,
$$\mathbf{n} = \overrightarrow{P_{0}P_{1}} \times \overrightarrow{P_{0}P} = \mathbf{u} \times \mathbf{v} = \begin{bmatrix}n_{1} \\ n_{2} \\ n_{3}\end{bmatrix}.$$
Let $\mathbf{r}_{0}$ be the position vector of the point $P_{0}$ and $\mathbf{r}$ the position vector of an arbitrary point on the plane $\Pi$. From the diagram, $\overrightarrow{P_{0}P}$ is perpendicular to $\mathbf{n}$. Therefore, from the definition of vector dot product,
$$\overrightarrow{P_{0}P} \cdot \mathbf{n} = 0 \iff (\overrightarrow{OP} - \overrightarrow{OP_{0}}) \cdot \mathbf{n} = 0.$$
Substituting $\overrightarrow{OP} = \mathbf{r}$ and $\overrightarrow{OP_{0}} = \mathbf{r}_{0}$ yields Equation $1$, as desired.


Corollary: Cartesian equation of a plane.
Let $\Pi$ be a plane in $\mathbb{R}^{3}$ as defined by Equation $1$.
Then the Cartesian equation of $\Pi$ is expressible as
$$ax + by + cz = d. \qquad \qquad \qquad (2)$$
In particular, $\Pi$ is an infinite set of points such that
$$\Pi = \left\{(x, y, z) \in \mathbb{R}^{3} \,\, \middle\vert \,\, ax + by + cz = d\right\}.$$

Proof:
Because
$$\overrightarrow{OP} = \mathbf{r}= \begin{bmatrix}x \\ y \\ z\end{bmatrix},$$
we get
$$\begin{alignedat}{3}
 && \mathbf{r}\cdot\mathbf{n} &= \mathbf{r}_{0}\cdot\mathbf{n} \\
 \iff && \begin{bmatrix}x \\ y \\ z\end{bmatrix}\cdot\begin{bmatrix}n_{1} \\ n_{2} \\ n_{3}\end{bmatrix} &= \begin{bmatrix}x_{0} \\ y_{0} \\ z_{0}\end{bmatrix}\cdot\begin{bmatrix}n_{1} \\ n_{2} \\ n_{3}\end{bmatrix} \\
 \iff && n_{1}x + n_{2}y + n_{3}z &= n_{1}x_{0} + n_{2}y_{0} + n_{3}z_{0}. \qquad \qquad \qquad (3)
\end{alignedat}$$
Let
$$\begin{bmatrix}n_{1} \\ n_{2} \\ n_{3}\end{bmatrix} = \begin{bmatrix}a \\ b \\ c\end{bmatrix}$$
and the righ-hand side of Equation $3$ be $d$, then the Cartesian equation of a plane in space is expressible as
$$ax + by + cz = d,$$
as desired.
