OK, of course I can solve a system with three unknowns, but here I have λ as well which is what, a coefficient?!
λx + 3y + z = 0
x + (λ + 1)y - z = 0
(2λ - 1)x + 2y + 4z = 0
The problem wants me to find values of λ where system has nontrivial solutions.
The system has solutions for all $\lambda$. If you are looking for unique solutions then you need the determinant of the matrix below to be non-zero:
$$ \left|\begin{matrix} \lambda&3&1\\1&\lambda+1&-1\\2\lambda-1&2&4 \end{matrix}\right|\ne0 $$
This implies that $2\lambda^2-\lambda-6=(2\lambda+3)(\lambda-2)\ne0$. So as long as $\lambda\ne2$ and $\lambda\ne\frac{-3}{2}$ you will have a unique solution to the system. This unique solution is $x=y=z=0$ which is trivial(you can get this by row-reducing the augmented matrix for the system).
If $\lambda=2$ or $\lambda=-\frac{3}{2}$ you will have non-trivial solutions because there will be an infinite family of solutions. For example when $\lambda=2$ you get solutions $y=z$, $x=-2z$ and $z$ can be whatever number you like. So you have the 'line' of solutions:
$$ z\cdot\left(\begin{matrix} -2\\1\\1 \end{matrix}\right) $$