1. The problem statement, all variables and given/known data Find the equation of the plane through the points on the x, y, and z-axes where x=a, y=b, z=c, respectively. 2. Relevant equations If <A,B,C> is a vector perpendicular to the plane, then A(x-a)+B(y-b)+C(z-c)=0 is the equation of the plane. 3. The attempt at a solution I know this is extremely simple, but I can't get the right answer even using different methods. Method 1: The vectors <0,b,-c> and <a,-b,0> are in the plane. The cross product <0,b,-c> X <a,-b,0> gives <-bc,-ac,-ab>, which is normal to the plane. Then the equation of the plane is (-bc)(x-a)+(-ac)(y-b)+(-ab)(z-c)=0, and simplifying I get (x/a)+(y/b)+(z/c)=3. Wrong. Method 2: Substituting the 3 points (a,0,0), (0,b,0), (0,0,c) into A(x-a)+B(y-b)+C(z-c)=0 gives -Bb-Cc=0 -Aa-Cc=0 -Aa-Bb=0. Thus C=-Aa/c, B=-Aa/b and substituting this back into A(x-a)+B(y-b)+C(z-c)=0 and simplifying I get (-x/a)+(y/b)+(z/c)=1, which is different than the answer from Method 1 and still wrong.