Find plane Π parallel to 5y-5x-5z=4 containing point p=(2,3,4)
Any plane parallel to 5y-5x-5z=4 has the form 5y-5x-5z=k, or on simplification, y-x-z=k.
Since p lies in the plane, the coordinates of p must satisfy y-x-z=k, or
k=3-2-4=-3
=>
π : y-x-z=-3
equivalently,
π : 5y-5x-5z=-15
