Give two linear functions ux y z ax by cz vx y z dx ey
     Give two linear functions: {u(x, y, z) = ax + by + cz  v(x, y, z) = dx + ey + fz  such that B = (-3  -3  1) is a basis for the vector space of solutions to the linear system  {u(x, y, z) = 0, v(x, y, z) =0}.   
  
  Solution
B = ( -3,-3,1)T is a basis for the solution space of the linear system u(x,y, z) = ax+by+cz = 0 and v(x,y,z) = dx+ey +fz = 0. Hence u(x,y,z) = -3px-3py+pz and v(x,y,z) = -3qx-3qy+qz where p and q are scalars. ( every solution to the given linear system has to be scalar multiple of the vectori the basis of the solution space).

