Find an equation of the plane I have come across a difficult problem in my textbook.
The problem ask:

Find an equation of the plane that passes through the line of intersection  of the planes $x-z=1$ and $y+2z=3$ and is perpendicular to the plane $x+y-2z=1$.

So far I found the direction vector of the line of intersection to be $<1,−2,1>$ and I have identified a point on this line when $x=0$ to be $(0,5,−1)$.
I do not know how to find the desired plane from here.
Any assistance would be appreciated.
 A: The plane should be parallel to the vector $a=<1,-2,1>$ and $b=<1,1,-2>$. Now we need a vector which is normal to these two vectors as a result a normal to the plane. Let $c=a \times b$ where $\times$ represent cross product. By definition of cross product, $c$ is normal to the plane. So, let $c$ be $<m,n,o>$. The equation of the plane is $$mx+n(y-5)+o(z+1)=0$$. 
A: First, you found the direction vector of the line, which is the intersection of the planes $x-z=1$ and $y+2z=3$, correctly:
$$
\vec{l}=(1,-2,1).
$$ 
You can take your point as
$$
M=(0,5,-1).
$$
You are also required that your plane is perpendicular to $x+y-2z=1$, which means that the normal to your sought plane $\vec{n}$ has to be orthogonal to $\vec{n}_1=(1,1,-2)$ (normal of the plane). So, you are looking for the plane passing through the point $M$ and such that its normal $\vec{n}\perp \vec{l}$ and $\vec{n}\perp \vec{n}_1$. To find such $\vec{n}$ you need to use the cross product $\vec{n}=\vec{l}\times\vec{n}_1$, or, in coordinates:
$$
\vec{n}=\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
1 &  -2 & 1 \\
1 &  1 & -2
\end{vmatrix}=3\mathbf{i}+3\mathbf{j}+3\mathbf{k},
$$
hence your vector is $\vec{n}=(1,1,1)$.
Finally, you find your plane:
$$
(x-0)+(y-5)+(z+1)=0,
$$
or
$$
x+y+z=4.
$$
A: A plane in Euclidean space is determined completely by the following:
1) A direction vector perpendicular to the plane, and
2) A single point on the plane.
You say you've found a point that is in the plane you wish to describe. Can you get a perpendicular vector? (Hint: look at the second condition.) And once you have 1) and 2), do you know how to write the equation of the plane determined by them?
A: A related problem. Already, you have got a point which lies in the plane. What's left is the normal to the plane. You are given that the plane is perpendicular to a plane which its normal is $n_1=(1,1,-2)$. To find the normal to the plane, assume 
$$ n=(a,b,c) \implies n.n_1=0 \implies (a,b,c).(1,1,-2)=0\implies a+b-2c=0. $$
Solve for $a,b$, and $c$ to get the normal vector. You will have infinite number of solutions, just pick up one. One possible solution is $n=(1,1,1)$.
