If the system of equations

$\begin{array}{l}x+y+z=2\\ 2x+4y-z=6\\ 3x+2y+\lambda z=\mu \end{array}$

Has infinitely many solutions, then:

$\begin{array}{l}x+y+z=2\\ 2x+4y-z=6\\ 3x+2y+\lambda z=\mu \end{array}$

For infinite solutions

$\begin{array}{l}\mathrm{D}=\left|\begin{array}{ccc}1& 1& 1\\ 2& 4& -1\\ 3& 2& \lambda \end{array}\right|=0\\ 1(4\lambda +2)-(2\lambda +3)+1(4-12)=0\\ 2\lambda -1-8=0 \lambda =\frac{9}{2}\\ {\mathrm{D}}_1=0={\mathrm{D}}_2={\mathrm{D}}_3\\ {\mathrm{D}}_1=\left|\begin{array}{lll}1& 1& 2\\ 2& 4& 6\\ 3& 2& \mu \end{array}\right|=0\\ 1(4\mu -12)-(2\mu -18)+2(4-12)=0\\ 4\mu -12-2\mu +18-16=0\\ 2\mu -10=0\\ \mu =5\\ \text{ now }2\lambda +\mu =9+5=14\end{array}$