Your book may say that but the problem, as you originally stated it, has nothing to do with either a "null space" nor a "row space". Both of those are properties of a linear transformation or matrix and you do not have either. You are given a subspace, W, of R3, given by W= {(x,y,z)| 2x-y+3z=0 }. If we take x= 1, z= 0, then 2- y= 0 so y= 2. (1, 2, 0) is in W. If we take x= 0, z= 1, then -y+ 3= 0 so y= 3. (0, 1, 3) is also in W and it should be easy to see that {(1, 2, 0), (0, 1, 3)} is a basis for W. The "orthogonal complement" of W consists of vectors that are perpendicular to all vectors in W and that is true if and only if a vector is perpendicular to both (1, 2, 0) and (0, 1, 3). Since two vectors in Rn are perpendicular if and only if their dot product is 0, (x, y, z) is in the orthogonal complement of W if and only if (1, 2, 0).(x, y, z)= x+ 2y= 0 and (0, 1, 3).(x, y, z)= y+ 3z= 0. What (x, y, z) satisfy both of those? Since those are two equations in 3 unknown values, you can solve for two of the, say x and z, in terms of the third, y, and write all vectors in the orthogonal complement of W in terms of y.