given the system of linear equations
$\begin{array}{rrrrr}x_{1} & + & x_{2} & + & 2 x_{3} & = & 2 \\ 2 x_{1} & + & 5 x_{2} & - & x_{3} & = & 2 \\ 3 x_{1} & + & 6 x_{2} & + & x_{3} & = & 4\end{array}$
I found the solutions
$x=\left(\begin{array}{c}\frac{8}{3}-\frac{11}{3} t \\ -\frac{2}{3}+\frac{5}{3} t \\ t\end{array}\right)$
Now I should find another solution $\mathbf{x} \neq \lambda \mathbf{y}$ for all $\lambda \in \mathbb{R}$. How am I supposed to do this?
I always like to see these things geometrically.
The solutions you've found give a parametrization of a line $\ell$ in $\Bbb{R}^3$ which does not pass through the origin.
Notice that two distinct solutions $\mathbf{x},\mathbf{y}$ will determine a unique line, namely $\ell$ itself.
The condition $\mathbf{x}\neq \lambda\mathbf{y}$ for all $\lambda$ means that $\mathbf{x},\mathbf{y}$ are linearly independent or, in other words, the line they determine does not pass through the origin.
What does that tell us?