If the system of equations
$$x + y + z = 2$$
$$2x + 4y - z = 6$$
$$3x + 2y + \lambda z = \mu$$
has infinitely many solutions, then:

We are given the three linear equations

$$x+y+z=2,$$

$$2x+4y-z=6,$$

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

For a system of three equations in the three variables $$x,\;y,\;z$$ to possess infinitely many solutions, the following well-known condition from the theory of linear equations must hold: the rank of the coefficient matrix and the rank of the augmented matrix must be equal, and both must be strictly less than the number of variables. Concretely, this means that one of the three equations must be expressible as a linear combination of the other two. In other words, there must exist real numbers $$a$$ and $$b$$ such that

$$a\,(x+y+z=2)+b\,(2x+4y-z=6)\;=\;(3x+2y+\lambda z=\mu).$$

We now equate coefficients term by term. Starting with the $$x$$-coefficients we have

$$a\cdot1\;+\;b\cdot2\;=\;3.$$

For the $$y$$-coefficients we have

$$a\cdot1\;+\;b\cdot4\;=\;2.$$

For the $$z$$-coefficients we have

$$a\cdot1\;+\;b\cdot(-1)\;=\;\lambda.$$

Finally, equating the constants on the right-hand side we get

$$a\cdot2\;+\;b\cdot6\;=\;\mu.$$

We first solve for $$a$$ and $$b$$ by using the two equations that do not involve $$\lambda$$ or $$\mu$$:

$$\begin{aligned} a+2b&=3,\\ a+4b&=2. \end{aligned}$$

Subtracting the first equation from the second gives

$$\bigl(a+4b\bigr)-\bigl(a+2b\bigr)=2-3\quad\Rightarrow\quad2b=-1\quad\Rightarrow\quad b=-\dfrac12.$$

Substituting $$b=-\dfrac12$$ into $$a+2b=3$$ yields

$$a+2\!\left(-\dfrac12\right)=3\quad\Rightarrow\quad a-1=3\quad\Rightarrow\quad a=4.$$

We now find $$\lambda$$ using $$a-b=\lambda$$ (from the $$z$$-coefficients):

$$\lambda=a-b=4-\left(-\dfrac12\right)=4+\dfrac12=\dfrac92.$$

Next we calculate $$\mu$$ using $$2a+6b=\mu$$ (from the constants):

$$\mu=2\cdot4+6\!\left(-\dfrac12\right)=8-3=5.$$

Thus, for infinitely many solutions, we must have

$$\lambda=\dfrac92,\qquad \mu=5.$$

We now test each option:

A. $$\lambda+2\mu=\dfrac92+2\cdot5=\dfrac92+10=\dfrac{29}2\neq14.$$

B. $$2\lambda-\mu=2\!\left(\dfrac92\right)-5=9-5=4\neq5.$$

C. $$\lambda-2\mu=\dfrac92-10=\dfrac92-\dfrac{20}2=-\dfrac{11}2\neq-5.$$

D. $$2\lambda+\mu=2\!\left(\dfrac92\right)+5=9+5=14,$$ which is true.

Hence, the correct answer is Option D.