TITLE: Find a plane with distance $3$ from $3x-y-z = 0$
QUESTION [5 upvotes]: I need to find a plane such that its distance from the plane $3x-y-z = 0$ is $3$. Since distance is defined only for parallel planes, I already know that they have to be parallel, and then, the equation of the new plane will have the same normal vector $(3,-1,-1)$.
Also, this plane intersects the origin, because $(0,0,0)$ satisfies its equation. So my idea was to normalize the normal vector, and then multiply it by $3$. I could then put it in the origin and see it as a point of distance $3$ from the origin (and also from the plane), and then this point should be in the new plane, so it must satisfy is equation, which is:
$$3x-y-z + d = 0$$
I think this migth work, but I don't think it's the best way to solve this exercise.
A friend of mine sent me a solution like this:
$$3(x-x_0) -1(y-y_0) -1(z-z_0) = 0 \\3x -y -z + (-3x_0 +y_0+z_0) = 0$$
then we should find $P = (x_0,y_0,z_0)$ such that $-3x_0 +y_0+z_0 = 0$. Then, $3P$ should be a point of the new plane, and therefore satisfy its new equation.
What's the best way to solve this, and could you explain my what my friend did in his solution?

REPLY [3 votes]: I decided to add the comment I made as a solution. I think this is a bit more intuitive for me. The solution posed by Harish is nice as well and I mean nothing by the statement "would it not be easier" since its all about personal preference.