Hi

The three variables do not exactly cancel out

(1;b;c)x(1;2;3)= t(5;-10;5)

gives

5t = 3b-2c

-10t = c-3

5t = 2-b

from which you can find one relation between b and c which is c = 2b-1

You could also find this relation by the dot product (1;b;c).(5;-10;5)=0

Actually you have found one direction vector of the common line but this is not sufficient since all parallel lines have the same direction vector

You should find one point of the common line (using the equation of P1 and P2)

Then using the fact that this point is also on P3 will give you a second relation between b and c

Solve the linear system for b and c (I can find b=3 and c=5)