Note that if \(\displaystyle{\left({x},{y},{z}\right)}∈{P}{1}\) then \(\displaystyle{2}{x}+{3}{y}−{z}={0}.\) Therefore we have \(\displaystyle{z}={2}{x}+{3}{y}\)

\(\displaystyle{\left({x},{y},{z}\right)}={\left({x},{y},{2}{x}+{3}{y}\right)}={\left({1},{0},{2}\right)}{x}+{\left({0},{1},{3}\right)}{y}∈{S}{p}{a}{n}{\left\lbrace{\left({1},{0},{2}\right)},{\left({0},{1},{3}\right)}\right\rbrace}\)

It follows that the vectors (1,0,2), (0,1,3) spans the plane P1.

\(\displaystyle{\left({x},{y},{z}\right)}={\left({x},{y},{2}{x}+{3}{y}\right)}={\left({1},{0},{2}\right)}{x}+{\left({0},{1},{3}\right)}{y}∈{S}{p}{a}{n}{\left\lbrace{\left({1},{0},{2}\right)},{\left({0},{1},{3}\right)}\right\rbrace}\)

It follows that the vectors (1,0,2), (0,1,3) spans the plane P1.