Take vector \(\displaystyle{n}={\left({x},{y},{z}\right)}\). As you said by yourself, dot product should vanish. So

\(\displaystyle{\left({4},{7},-{9}\right)}\cdot{\left({x},{y},{z}\right)}={4}{x}+{7}{y}-{9}{z}={0}\)

As you might see, all points that lie on the plane \(\displaystyle{4}{x}+{7}{y}-{9}{z}={0}\) satisfy the condition of perpendicularity. If you want two linear independent vectors, just pick two different points. So, pick any two different triples \(\displaystyle{n}_{{1}}={\left({x}_{{1}},{y}_{{1}},\frac{{{4}{x}_{{1}}+{7}{y}_{{1}}}}{{9}}\right)}\) and \(\displaystyle{n}_{{2}}={\left({x}_{{2}},{y}_{{2}},\frac{{{4}{x}_{{2}}+{7}{y}_{{2}}}}{{9}}\right)}\) where \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}\ne{\left({x}_{{2}},{y}_{{2}}\right)}\), and you're done.

\(\displaystyle{\left({4},{7},-{9}\right)}\cdot{\left({x},{y},{z}\right)}={4}{x}+{7}{y}-{9}{z}={0}\)

As you might see, all points that lie on the plane \(\displaystyle{4}{x}+{7}{y}-{9}{z}={0}\) satisfy the condition of perpendicularity. If you want two linear independent vectors, just pick two different points. So, pick any two different triples \(\displaystyle{n}_{{1}}={\left({x}_{{1}},{y}_{{1}},\frac{{{4}{x}_{{1}}+{7}{y}_{{1}}}}{{9}}\right)}\) and \(\displaystyle{n}_{{2}}={\left({x}_{{2}},{y}_{{2}},\frac{{{4}{x}_{{2}}+{7}{y}_{{2}}}}{{9}}\right)}\) where \(\displaystyle{\left({x}_{{1}},{y}_{{1}}\right)}\ne{\left({x}_{{2}},{y}_{{2}}\right)}\), and you're done.