Intersection of Three Planes

You will need to find expressions for the three planes in the standard vector form r.n = p, where r is a general point on the plane, n is a vector normal to the plane and p is a scalar constant (different for each plane).

For P1 you can reason that n is perpendicular to AC and BC since those two vectors are parallel to the plane. You can use any of the three known points in the plane to fix p.

P2 and P3 are a bit easier as you are effectively given the two normal vectors.

Once you have vector expressions for the three planes you will need to find the line intersection of two of them and then find where that line cuts the third plane.

..

For P1 I got -20y+10z+40=0
P2: 4x-14y+17z+40=0
P3: 4x+6y+13z-40=0

Don't really know how to solve this

I think there may be a slightly better way here:

We can write a general point r on $P_1$ in the form ${\bf r} = {\bf a} + \lambda ({\bf b-a}) + \mu ({\bf c-a})$

We can more or less write down conditions (of the form r.x = K for suitable vector x and constant K) for the other two planes. Sub in the general point for P_1 and you should end up with 2 simultaneous equations for $\lambda, \mu$. Solve that, and finally calculate r.