Let the system of linear equations
$$4x + \lambda y + 2z = 0$$
$$2x - y + z = 0$$
$$\mu x + 2y + 3z = 0$$, $$\lambda, \mu \in R$$
has a non-trivial solution. Then which of the following is true?

For the homogeneous system $$4x + \lambda y + 2z = 0$$, $$2x - y + z = 0$$, $$\mu x + 2y + 3z = 0$$ to have a non-trivial solution, the determinant of the coefficient matrix must be zero.

The determinant is $$\begin{vmatrix} 4 & \lambda & 2 \\ 2 & -1 & 1 \\ \mu & 2 & 3 \end{vmatrix}$$. Expanding along the first row: $$4(-3 - 2) - \lambda(6 - \mu) + 2(4 + \mu) = -20 - \lambda(6 - \mu) + 8 + 2\mu = -12 - 6\lambda + \lambda\mu + 2\mu$$.

Setting this to zero: $$\mu(\lambda + 2) - 6(\lambda + 2) = 0$$, which factors as $$(\mu - 6)(\lambda + 2) = 0$$. So either $$\mu = 6$$ or $$\lambda = -2$$. In particular, when $$\mu = 6$$ the determinant is zero for every value of $$\lambda$$, so the system has a non-trivial solution for all $$\lambda \in \mathbb{R}$$.