I have to show the locus of the points $P$ from a distance $d$ of a plane $\alpha$ are the union of two parallel planes to $\alpha$.
What I did: since $$\alpha\colon ax+by+cz+\tilde{d}=0$$ then the normal vector of $\alpha$ is $\vec{n}=(a,b,c)$. Also, I know $$d_{\alpha,P}=\frac{|ax+by+cz+\tilde{d}|}{\underbrace{\sqrt{a^2+b^2+c^2}}_{=K}}=d\Rightarrow|ax+by+cz+\tilde{d}|=\underbrace{K\cdot d}_{=M}$$
Hence $$ax+by+cz+\tilde{d}=M\Rightarrow ax+by+cz+(\tilde{d}-M)=0$$ or $$ax+by+cz+\tilde{d}=-M\Rightarrow ax+by+cz+(\tilde{d}+M)=0$$ which are both equations of planes with normal $(a,b,c)$, therefore, parallels to $\alpha$.
Is it correct?