Thread: Intersection line of planes, vector geometry problem

Suppose planes P1: 4x+3y+2z+1=0 and P2: x+2y+3z+4=0 and P3: x+by+cz+7=0 possess a common line of intersection. Find b and c.
My approach was first to perform the cross product of the normal vectors of p1 and p2. The result is the direction vector of the line of intersection of all three planes. The direction vector is (5;-10;5). Thus the cross product of P1 or P2 and P3, must have as result t(5;-10;5), too. Therefore: (1;b;c)x(1;2;3)= t(5;-10;5). Unfortunately, all the three variables (a,b,t) cancel each other out.
Where is my mistake? How to solve this problem?

Find a point on two planes.
is on .
It must also be on .
Work from there.