iii) \( \left[\begin{array}{c}2 x+2 y+2 \omega=0 \\ x+y+\omega=0 \\ 3 x+3 y+3 \omega=0\end{array}\right. \).

To solve the system of equations given by \[ \begin{cases} 2x + 2y + 2\omega = 0 \\ x + y + \omega = 0 \\ 3x + 3y + 3\omega = 0 \end{cases} \] we can simplify the equations. Notice that the first and third equations can be simplified by dividing through by their respective coefficients. 1. The first equation can be simplified by dividing by 2: \[ x + y + \omega = 0 \] 2. The second equation is already in its simplest form: \[ x + y + \omega = 0 \] 3. The third equation can be simplified by dividing by 3: \[ x + y + \omega = 0 \] Now we see that all three equations are actually the same: \[ x + y + \omega = 0 \] This means that we have a single equation with three variables. We can express one variable in terms of the others. For example, we can express \(\omega\) in terms of \(x\) and \(y\): \[ \omega = -x - y \] Thus, the solution set can be expressed in terms of two free variables, \(x\) and \(y\). We can write the general solution as: \[ \begin{pmatrix} x \\ y \\ \omega \end{pmatrix} = \begin{pmatrix} x \\ y \\ -x - y \end{pmatrix} \] where \(x\) and \(y\) can take any real values. In conclusion, the solution set is: \[ \{ (x, y, -x - y) \mid x, y \in \mathbb{R} \} \] This represents a plane in \(\mathbb{R}^3\).