The system of equations can be written in the matrix form as

$$\left[ {\matrix{ 2 & 4 & { - \lambda } \cr 4 & \lambda & 2 \cr \lambda & 2 & 2 \cr } } \right]\left[ {\matrix{ x \cr y \cr z \cr } } \right] = \left[ {\matrix{ 0 \cr 0 \cr 0 \cr } } \right]$$

The system has infinite solutions; thus, we get

$$\left| {\matrix{ 2 & 4 & { - \lambda } \cr 4 & \lambda & 2 \cr \lambda & 2 & 2 \cr } } \right| = 0$$

$$ \Rightarrow 0 = 2(2\lambda - 4) - 4(8 - 2\lambda ) - \lambda (8 - {\lambda ^2})$$

$$ \Rightarrow 4\lambda - 8 - 32 + 8\lambda - 8\lambda + {\lambda ^3} = 0$$

$$ \Rightarrow {\lambda ^3} + 4\lambda - 40 = 0$$

We can solve this by graphical method:

$$y = {x^3} + 4x - 40$$

For x = 0, y = $$-$$40: If we take y = $$-$$40, then we have

$$ - 40 = {x^3} + 4x - 40$$

$$ \Rightarrow {x^3} + 4x = 0$$

$$ \Rightarrow x({x^2} + 4) = 0$$

$$ \Rightarrow x = 0,{x^2} + 4 = 0$$

$$ \Rightarrow x = \pm \,2i$$

The given equation of line intersects x only at one point; therefore, the real value of $$\lambda$$ is only one.