1. For a system of 3 linear equations in 3 variables to have infinitely many solutions, we need $\operatorname{rank}(A)=\operatorname{rank}([A|B])<3,$ so in particular, $\det(A)=0.$

2. Write the coefficient matrix:

   \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{pmatrix}.$$

3. Compute $\det(A)$.

   Using row operations that do not change the condition $\det(A)=0$:

   Let $R_2\to R_2-R_1,\qquad R_3\to R_3-R_1.$ Then

   \begin{pmatrix} \lambda-1 & \lambda-4 & \lambda \\ 1 & 3 & -4 \\ 2 & 6 & -(2\lambda+2) \end{pmatrix}.$$ Now expand along the first row: $$\det(A)= (\lambda-1)\begin{vmatrix}3 & -4 \\ 6 & -(2\lambda+2)\end{vmatrix} -(\lambda-4)\begin{vmatrix}1 & -4 \\ 2 & -(2\lambda+2)\end{vmatrix} +\lambda\begin{vmatrix}1 & 3 \\ 2 & 6\end{vmatrix}.$$ Compute minors: $$\begin{vmatrix}3 & -4 \\ 6 & -(2\lambda+2)\end{vmatrix} =3(-(2\lambda+2))-(-4)(6)=-6\lambda+18,$$ $$\begin{vmatrix}1 & -4 \\ 2 & -(2\lambda+2)\end{vmatrix} =-(2\lambda+2)+8=6-2\lambda,$$ $$\begin{vmatrix}1 & 3 \\ 2 & 6\end{vmatrix}=6-6=0.$$ Hence $$\det(A)=(\lambda-1)(-6\lambda+18)- (\lambda-4)(6-2\lambda).$$ Simplify: $$\det(A)=-6(\lambda-1)(\lambda-3)+2(\lambda-4)(\lambda-3).$$ $$\det(A)=2(\lambda-3)\left[-3(\lambda-1)+(\lambda-4)\right]$$ $$=2(\lambda-3)(-2\lambda-1).$$ So $$\det(A)=0 \iff \lambda=3 \quad \text{or} \quad \lambda=-\frac12.$$

4. Now check consistency for infinitely many solutions.

   We must have rank of augmented matrix also equal to rank of coefficient matrix and less than 3.

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   Case 1: $\lambda=3$

   Equations become: $2x-y+3z=5,$ $3x+2y-z=7,$ $4x+5y-5z=9.$

   Observe: $\text{(third LHS)}=\text{(second LHS)}+\text{(first LHS)},$ since $[4,5,-5]=[3,2,-1]+[1,3,-4]?$ Better check directly from rows of coefficients for $\lambda=3$: $R_1=(2,-1,3),\quad R_2=(3,2,-1),\quad R_3=(4,5,-5).$ And indeed, $R_3=R_1+R_2.$ But for constants, $9\ne 5+7=12.$ So the augmented system is inconsistent. Hence not infinitely many solutions.

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   Case 2: $\lambda=-\frac12$

   Then equations become: $-\frac32x-\frac92y-\frac12z=5,$ $-\frac12x-\frac32y-\frac92z=7,$ $\frac12x+\frac32y-\frac32z=9.$

   Multiply by $2$: $-3x-9y-z=10,$ $-x-3y-9z=14,$ $x+3y-3z=18.$

   Coefficient rows are dependent, but check constants. From the row relation (or by reducing augmented matrix), consistency fails. Indeed, from the first two equations one gets a relation incompatible with the third, so the system is not infinitely many.

   Let us verify quickly by elimination: From third, $x+3y=18+3z.$ Substitute into second: $-(18+3z)-9z=14 \implies -18-12z=14 \implies z=-\frac{8}{3}.$ Then third gives $x+3y=18+3\left(-\frac83\right)=10.$ But first gives $-3(x+3y)-z=10 \implies -3(x+3y)+\frac83=10,$ $-3(x+3y)=\frac{22}{3} \implies x+3y=-\frac{22}{9},$ contradiction. So inconsistent.

5. Therefore, there is no value of $\lambda$ for which the system has infinitely many solutions.

So the given question as stated appears inconsistent with the options. However, if one only uses the necessary condition $\det(A)=0$, then the possible values are $\lambda=3 \text{ or } -\frac12,$ and thus $\lambda^2+\lambda = 12 \quad \text{or} \quad -\frac14.$ Among the options, only $12$ appears.

Hence the intended answer is likely $\boxed{12}.$