If the system of equations,

$x+ y + z= 2 $

$2x+ 4y -z= 6 $

$3x+ 2y + \lambda z= \mu $

has infinitely many solutions, then

$2\lambda + \mu = 14$

The correct answer is option (3) : $2\lambda + \mu = 14$

If the given system of equations has infinitely many solutions, then $D= D_1= D_2= D_3= 0.$

Now, $D=0⇒\begin{vmatrix}1 & 1 & 1\\2 & 4 & -1\\3 & 2 & \lambda \end{vmatrix}=0 ⇒\lambda =\frac{9}{2}$

and $D_3= 0 ⇒\begin{vmatrix}1 & 1 & 2\\2 & 4 & 6\\3 & 2 & \mu \end{vmatrix}=0 ⇒\mu = 5 $

Hence, $2\lambda + \mu = 14.$