The system of equations

$\lambda x+ 2y + 2z= 5$

$2\lambda + 3y + 5z= 8 $

$4x+ \lambda y + 6z= 10 $

has

no solution when $\lambda = 2$

The correct answer is option (4) : no solution when $\lambda = 2$

For the given system of equations

$\begin{vmatrix}\lambda & 2 & 2\\2\lambda & 3 & 5\\4 & \lambda  & 6\end{vmatrix}= ( \lambda  +8) (2 - \lambda ), D_1 = \begin{vmatrix}5 & 2 & 2\\8 & 3 & 5\\10 & \lambda  & 6\end{vmatrix}= 34- 9 \lambda$

Clearly, D =0 for $\lambda = 2, -8 $ but $D_1≠0$ for these values of $\lambda .$

Hence, the system of equations has no solution for $\lambda =2, -8.$