Find a combination$~x_{1}\mathbf{\omega}_{1}+x_2\mathbf{\omega}_{2}+x_{3}\mathbf{\omega}_{3}~$that gives the zero vector with$~x_1=1:~$ $$\mathbf{\omega}_{1}= \begin{bmatrix} 1\\2\\3 \end{bmatrix}~~~~ \mathbf{\omega}_{2}= \begin{bmatrix} 4\\5\\6 \end{bmatrix}~~~~ \mathbf{\omega}_{3}= \begin{bmatrix} 7\\8\\9 \end{bmatrix} $$ Those vectors are(independent)(dependent). The three vectors lie in a___. The matrix$~W~$with those three columns is not invertible.
Firtst things to first, as set of$~x_1=1,x_2,x_3~\text{exists}~\operatorname{s. t. }~x_{1}\mathbf{\omega}_{1}+x_2\mathbf{\omega}_{2}+x_{3}\mathbf{\omega}_{3}=\mathbf 0~$,set of vectors$~\mathbf{\omega}_{1},\mathbf{\omega}_{2},\mathbf{\omega}_{3}~$is dependent from the definition of linear dependence. So vectors$~\mathbf{\omega}_{1},\mathbf{\omega}_{2},\mathbf{\omega}_{3}~~$lie in a plane.
$$x_1\mathbf{\omega_{1}}+x_2\mathbf{\omega_{2}}+x_2\mathbf{\omega_{2}}=\mathbf{0}_{}\tag{1}$$
$$1\cdot\mathbf{\omega_{1}}+x_2\mathbf{\omega_{2}}+x_2\mathbf{\omega_{2}}=\mathbf{0}_{}\tag{2}$$
$$\begin{pmatrix}1\\2\\3\end{pmatrix}+x_{2}\begin{pmatrix}4\\5\\6\end{pmatrix}+x_{3}\begin{pmatrix}7\\8\\9\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\tag{3}$$
$$x_{2}\begin{pmatrix}4\\5\\6\end{pmatrix}+x_{3}\begin{pmatrix}7\\8\\9\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}-\begin{pmatrix}1\\2\\3\end{pmatrix}\tag{4}$$
$$x_{2}\begin{pmatrix}4\\5\\6\end{pmatrix}+x_{3}\begin{pmatrix}7\\8\\9\end{pmatrix}=\begin{pmatrix}-1\\-2\\-3\end{pmatrix}\tag{5}$$
$$~~\text{Stucked here}~~\tag{6}$$
The combination$~0\mathbf{\omega}_{1}+0\mathbf{\omega}_{2}+0\mathbf{\omega}_{3}~$always gives the zero vector, but this problem looks for other zero combinations(then the vectors are dependent,they lie in a place): $~\mathbf{\omega}_{2}={\left(\mathbf{\omega}_{1}+\mathbf{\omega}_{3}\right)\over2}~$so one combinations that gives zero is$~\mathbf{\omega}_{1}-2\mathbf{\omega}_{2}+\mathbf{\omega}_{3}~$
$$~ \color{blue}{\underbrace{\mathbf{\omega}_{2}={\left(\mathbf{\omega}_{1}+\mathbf{\omega}_{3}\right)\over2},~\mathbf{\omega}_{1}-2\mathbf{\omega}_{2}+\mathbf{\omega}_{3}}_{\text{How these are obtained?}}} ~\tag{7}$$
The official answer states that"one combinations that"so I interpret that there might over one solutions exist. Is this thought correct?