If the system of linear equations
$$2x + 3y - z = -2$$
$$x + y + z = 4$$
$$x - y + |\lambda|z = 4\lambda - 4$$ where $$\lambda \in \mathbb{R}$$,
has no solution, then

Given system of equations,

$$2x+3y-z=-2$$

$$x+y+z=4$$

$$x-y+|\lambda|z=4\lambda-4$$

For the system to have no solution,

$$\Delta=0$$

and at least one of

$$\Delta_x,\Delta_y,\Delta_z$$

must be non-zero.

Now,

$$\Delta= \begin{vmatrix} 2&3&-1\\ 1&1&1\\ 1&-1&|\lambda| \end{vmatrix} $$

Expanding,

$$\Delta = 2(|\lambda|+1)-3(|\lambda|-1)+2 $$

$$=2|\lambda|+2-3|\lambda|+3+2$$

$$=7-|\lambda|$$

For no solution,

$$7-|\lambda|=0$$

$$|\lambda|=7$$

Hence,

$$\lambda=7\quad \text{or}\quad \lambda=-7$$

Now,

$$\Delta_z= \begin{vmatrix} 2&3&-2\\ 1&1&4\\ 1&-1&4\lambda-4 \end{vmatrix} $$

Expanding,

$$\Delta_z=28-4\lambda$$

For

$$\lambda=7$$

$$\Delta_z=28-28=0$$

Hence, the system is consistent.

For

$$\lambda=-7$$

$$\Delta_z=28+28=56\ne0$$

Thus,

$$\Delta=0\quad \text{and}\quad \Delta_z\ne0$$

Therefore, the system has no solution for

$$\boxed{\lambda=-7}$$

Hence, the correct answer is

$$\boxed{\text{Option B}}$$