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?
Equation of the plane parallel to the given plane: $3x-y-z=0$ is $$3x-y-z+c=0 $$ Where $c$ is an arbitrary constant. Now, using $\color{blue}{\text{distance formula for parallel planes}}$ as follows $$\frac{\left|c-0\right|}{\sqrt{(3)^2+(-1)^2+(-1)^2}}=3$$ $$\left|c\right|=3\sqrt{11}$$ $$c=\pm 3\sqrt{11}$$ Hence, there are two parallel planes at distance $3$ on either side to the given plane: $3x-y-z=0$. Hence, the equations of unknown planes $$\color{blue}{3x-y-z\pm3\sqrt{11}=0}$$
Edit: In general, the distance between any two parallel planes: $ax+by+cz+d_1=0$ & $ax+by+cz+d_2=0$ is $$=\frac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$$