To describe the system of equation in the parametric vector form, solve for \(\displaystyle{x}_{{1}},\ {x}_{{2}}\) and \(\displaystyle{x}_{{3}}\).

\(\displaystyle{3}{x}_{{1}}+{3}{x}_{{2}}+{6}{x}_{{3}}={0}\)

\(\displaystyle-{9}{x}_{{1}}-{9}{x}_{{2}}-{18}{x}_{{3}}={0}\)

\(\displaystyle-{6}{x}_{{2}}-{6}{x}_{{3}}={0}\)

Let Solving equation we get:

\(\displaystyle-{6}{x}_{{2}}={6}{x}_{{3}}\)

\(\displaystyle{x}_{{2}}=-{x}_{{3}}\)

Substitute \(\displaystyle{x}_{{2}}=-{x}_{{3}}\) in equation, we get:

\(\displaystyle{3}{x}_{{1}}-{3}{x}_{{3}}+{6}{x}_{{3}}={0}\)

\(\displaystyle{3}{x}_{{1}}=-{3}{x}_{{3}}\)

\(\displaystyle{x}_{{1}}=-{x}_{{3}}\)

Thus, the paranetric vector form for system of equation is as follows:

\[x=\begin{bmatrix}-1\\-1\\1\end{bmatrix}\]

\(\displaystyle{3}{x}_{{1}}+{3}{x}_{{2}}+{6}{x}_{{3}}={0}\)

\(\displaystyle-{9}{x}_{{1}}-{9}{x}_{{2}}-{18}{x}_{{3}}={0}\)

\(\displaystyle-{6}{x}_{{2}}-{6}{x}_{{3}}={0}\)

Let Solving equation we get:

\(\displaystyle-{6}{x}_{{2}}={6}{x}_{{3}}\)

\(\displaystyle{x}_{{2}}=-{x}_{{3}}\)

Substitute \(\displaystyle{x}_{{2}}=-{x}_{{3}}\) in equation, we get:

\(\displaystyle{3}{x}_{{1}}-{3}{x}_{{3}}+{6}{x}_{{3}}={0}\)

\(\displaystyle{3}{x}_{{1}}=-{3}{x}_{{3}}\)

\(\displaystyle{x}_{{1}}=-{x}_{{3}}\)

Thus, the paranetric vector form for system of equation is as follows:

\[x=\begin{bmatrix}-1\\-1\\1\end{bmatrix}\]