b
Since the given system of linear equations has infinitely many solutions.
$\therefore \begin{array}{*{20}{c}}
2&4&{ - \lambda }\\
4&\lambda &2\\
\lambda &2&2
\end{array} = 0$
$ \Rightarrow {\lambda ^3} + 4\lambda - 40 = 0$
$\lambda $ has only $1$ real root.