The given polynomials are

\(\displaystyle{2}{x}^{{3}}−{2}{x}^{{2}}+{12}{x}−{6}\)

\(\displaystyle{x}^{{3}}−{2}{x}^{{2}}−{5}{x}−{3}\)

\(\displaystyle{3}{x}^{{3}}−{5}{x}^{{2}}−{4}{x}−{9}\)

The polynomial can be written as vector form

\(v_1=\begin{bmatrix}2\\-2\\12\\-6\end{bmatrix},v_2=\begin{bmatrix}1\\-2\\-5\\-3\end{bmatrix},v_2=\begin{bmatrix}3\\-5\\-4\\-9\end{bmatrix}\)

Let, \(\displaystyle{v}_{{1}}={a}{v}_{{2}}+{b}{v}_{{3}}\)

\(\begin{bmatrix}2\\-2\\12\\-6\end{bmatrix}=a\begin{bmatrix}1\\-2\\-5\\-3\end{bmatrix}+b\begin{bmatrix}3\\-5\\-4\\-9\end{bmatrix}\)

\(\begin{bmatrix}2\\-2\\12\\-6\end{bmatrix}=\begin{bmatrix}a+3b\\-2a-5b\\5a-4b\\-3a-9b\end{bmatrix}\)

\(\displaystyle\Rightarrow{a}+{3}{b}={2}\Rightarrow-{5}{a}-{4}{b}={12}\)

\(\displaystyle\Rightarrow-{2}{a}-{5}{b}=-{2}\Rightarrow-{3}{a}-{9}{b}=-{6}\)

using equation

\(\displaystyle{6}{b}-{5}{b}={4}-{2}\)

b=2

a=-4

These values of a,b are also santisty, hence polynomial having linear relationship

\(\displaystyle{P}_{{1}}=-{4}{P}_{{2}}+{2}{P}_{{3}}\)

or \(\displaystyle{2}{P}_{{3}}={P}_{{1}}+{4}{P}_{{2}}\)

\(\displaystyle{P}_{{1}}=\) I polynomial

\(\displaystyle{P}_{{2}}=\) II polynomial

\(\displaystyle{P}_{{3}}=\) III polynomial