I have to work with Bisector planes. Can u pls help me?

Question

The points are A (1.0, -1) B (2.3.1) C (0.2, -3) Determine point P that’s from the same distance from A, B and C and at a distance √5 from the plane ABC. ![So this is how I solved it , but I have to do something with Bisector planes) (https://i.stack.imgur.com/FqKCt.jpg)

Answer 1

I would consider the following strategy:

1. Determine the normal vector n to the plane (Calculate (A-B) x (C-B)). Multiply by $\frac{\sqrt5}{|n|}$.

2. Find a point $\tilde{P}$ on the plane ABC that has equal distance from A,B and C. You can do so by intersecting the medians to $\overline{AB}$ and $\overline{BC}$.

3. $P=\tilde{P}+n\frac{\sqrt5}{|n|}$