find all vectors in R3 perpendicular to vector 1,3,-3

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Shandra, that's a tough question because there are so many perpendicular vectors.  Let's start by assuming that the perpendicular vectors have the form <x,y,z>.  We know that if two vectors are perpendicular, then their dot product will be zero.  That is, we can say that <1,3,-3> • <x,y,z> = 0.  Now expand that dot product and you'll get x+3y-3z=0.  That's the equation of a plane!  Every vector contained in this plane is perpendicular to <1,3,-3>.  For example, the point (3,0,1) in the plane corresponds to the vector <3,0,1>.  Observe that this vector is perpendicular to <1,3,-3>:<3,0,1> • <1,3,-3> = 3+0-3 = 0