I've got a question that I can't seem to figure out, any hint/help would be appreciated.
Be $L$ a line, and $P_1$ a point.
So, I've got $P_1$, and $L$, I know the distance from $P_1$ and $L$, let it be $c$, so I need to find the equation of a plane that has $P_1$ and also the distance is $c$.
My take on it is to find a plane that is to find the perpendicular vector of $L$ in respect of $P_1$, let that be the normal of the plane, and carry from there.
Assuming I understand the problem correctly by the discussion we had in the comments:
Let $\mathbf{x_0}$ and $\mathbf{x_1}$ be two points on your line $L$. You can express the shortest distance between any point $\mathbf{P}$ and this line by:
$$ d = \frac{|(\mathbf{x}_1 - \mathbf{x}_2) \times (\mathbf{x}_1 - \mathbf{P})|}{|\mathbf{x}_1 - \mathbf{x}_2|} $$
By using $\mathbf{P}_1\mathbf{P} = (p_1 - x, p_2 - y, p_3 - z)$ as your point, and setting $d$ equal to the distance $c$, you have an equation for a plane satisfying your problem.