If the system of equations $$(\lambda-1)x+(\lambda-4)y+\lambda z=5 \\\lambda x+(\lambda-1)y+(\lambda-4)z=7 \\ (\lambda+1)x+(\lambda+2)y-(\lambda+2)z=9$$ has infinitely many solutions, then $$\lambda^{2}+\lambda$$ is equal to

For the system of equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero, and the system must be consistent.

The coefficient matrix is:

$$A = \begin{bmatrix} \lambda-1 & \lambda-4 & \lambda \\ \lambda & \lambda-1 & \lambda-4 \\ \lambda+1 & \lambda+2 & -(\lambda+2) \end{bmatrix}$$

The determinant of A is computed as:

$$\text{det}(A) = (\lambda-1) \begin{vmatrix} \lambda-1 & \lambda-4 \\ \lambda+2 & -(\lambda+2) \end{vmatrix} - (\lambda-4) \begin{vmatrix} \lambda & \lambda-4 \\ \lambda+1 & -(\lambda+2) \end{vmatrix} + \lambda \begin{vmatrix} \lambda & \lambda-1 \\ \lambda+1 & \lambda+2 \end{vmatrix}$$

Calculate each 2x2 determinant:

First minor: $$\begin{vmatrix} \lambda-1 & \lambda-4 \\ \lambda+2 & -(\lambda+2) \end{vmatrix} = (\lambda-1)(-(\lambda+2)) - (\lambda-4)(\lambda+2) = -(\lambda+2)(2\lambda-5)$$

So, the first term is: $$(\lambda-1) \cdot [-(\lambda+2)(2\lambda-5)] = -(\lambda-1)(\lambda+2)(2\lambda-5)$$

Second minor: $$\begin{vmatrix} \lambda & \lambda-4 \\ \lambda+1 & -(\lambda+2) \end{vmatrix} = \lambda \cdot [-(\lambda+2)] - (\lambda-4)(\lambda+1) = -2\lambda^2 + \lambda + 4$$

So, the second term is: $$- (\lambda-4) \cdot (-2\lambda^2 + \lambda + 4) = -[-2\lambda^3 + 9\lambda^2 - 16] = 2\lambda^3 - 9\lambda^2 + 16$$

Third minor: $$\begin{vmatrix} \lambda & \lambda-1 \\ \lambda+1 & \lambda+2 \end{vmatrix} = \lambda(\lambda+2) - (\lambda-1)(\lambda+1) = 2\lambda + 1$$

So, the third term is: $$\lambda \cdot (2\lambda + 1) = 2\lambda^2 + \lambda$$

Summing all terms: $$\text{det}(A) = -(\lambda-1)(\lambda+2)(2\lambda-5) + 2\lambda^3 - 9\lambda^2 + 16 + 2\lambda^2 + \lambda$$

Expanding $$-(\lambda-1)(\lambda+2)(2\lambda-5)$$: $$(\lambda-1)(\lambda+2) = \lambda^2 + \lambda - 2$$ $$(\lambda^2 + \lambda - 2)(2\lambda - 5) = 2\lambda^3 - 3\lambda^2 - 9\lambda + 10$$ So, $$-(\lambda-1)(\lambda+2)(2\lambda-5) = -2\lambda^3 + 3\lambda^2 + 9\lambda - 10$$

Adding the other terms: $$(-2\lambda^3 + 3\lambda^2 + 9\lambda - 10) + (2\lambda^3 - 9\lambda^2 + 16) + (2\lambda^2 + \lambda) = -4\lambda^2 + 10\lambda + 6$$

Thus, $$\text{det}(A) = -4\lambda^2 + 10\lambda + 6$$.

Set $$\text{det}(A) = 0$$: $$-4\lambda^2 + 10\lambda + 6 = 0$$ Multiply by $$-1$$: $$4\lambda^2 - 10\lambda - 6 = 0$$ Divide by 2: $$2\lambda^2 - 5\lambda - 3 = 0$$

Solve the quadratic equation: $$\lambda = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 2 \cdot (-3)}}{4} = \frac{5 \pm \sqrt{25 + 24}}{4} = \frac{5 \pm \sqrt{49}}{4} = \frac{5 \pm 7}{4}$$ So, $$\lambda = \frac{12}{4} = 3$$ or $$\lambda = \frac{-2}{4} = -\frac{1}{2}$$.

Now, check consistency for each value.

Case $$\lambda = 3$$:

The system becomes: $$2x - y + 3z = 5 \quad \text{(1)}$$ $$3x + 2y - z = 7 \quad \text{(2)}$$ $$4x + 5y - 5z = 9 \quad \text{(3)}$$

Subtract equation (1) from equation (2): $$(3x + 2y - z) - (2x - y + 3z) = 7 - 5 \implies x + 3y - 4z = 2 \quad \text{(4)}$$

Subtract equation (1) from equation (3): $$(4x + 5y - 5z) - (2x - y + 3z) = 9 - 5 \implies 2x + 6y - 8z = 4$$ Divide by 2: $$x + 3y - 4z = 2 \quad \text{(5)}$$

Equations (4) and (5) are identical. The system reduces to: $$2x - y + 3z = 5 \quad \text{and} \quad x + 3y - 4z = 2$$

Solving for $$x$$ and $$y$$ in terms of $$z$$: From equation (5): $$x = 2 - 3y + 4z$$. Substitute into equation (1): $$2(2 - 3y + 4z) - y + 3z = 5 \implies 4 - 6y + 8z - y + 3z = 5 \implies -7y + 11z = 1$$ So, $$y = \frac{11z - 1}{7}$$, and $$x = 2 - 3\left(\frac{11z - 1}{7}\right) + 4z$$. For any $$z$$, there is a solution, so infinitely many solutions.

Case $$\lambda = -\frac{1}{2}$$:

The system becomes: $$-3x - 9y - z = 10 \quad \text{(1)}$$ $$-x - 3y - 9z = 14 \quad \text{(2)}$$ $$x + 3y - 3z = 18 \quad \text{(3)}$$

Add equations (2) and (3): $$(-x - 3y - 9z) + (x + 3y - 3z) = 14 + 18 \implies -12z = 32 \implies z = -\frac{8}{3}$$

Substitute $$z = -\frac{8}{3}$$ into equation (3): $$x + 3y - 3\left(-\frac{8}{3}\right) = 18 \implies x + 3y + 8 = 18 \implies x + 3y = 10$$

Substitute $$z = -\frac{8}{3}$$ into equation (1): $$-3x - 9y - \left(-\frac{8}{3}\right) = 10 \implies -3x - 9y + \frac{8}{3} = 10$$ Multiply by 3: $$-9x - 27y + 8 = 30 \implies -9x - 27y = 22 \quad \text{(4)}$$

From $$x + 3y = 10$$, multiply by $$-9$$: $$-9x - 27y = -90$$. But equation (4) gives $$-9x - 27y = 22$$, contradiction. So inconsistent.

Thus, only $$\lambda = 3$$ gives infinitely many solutions. Now compute $$\lambda^2 + \lambda$$: $$3^2 + 3 = 9 + 3 = 12$$

The answer is 12.