1. **State the problem:** We are given a determinant equation involving $\lambda$: $$\begin{vmatrix} \lambda - 1 & \lambda - 4 & \lambda \\ \lambda & \lambda - 1 & \lambda - 4 \\ \lambda + 1 & \lambda + 2 & -(\lambda + 2) \end{vmatrix} = 0$$ and the factorized form: $$(\lambda - 3)(2\lambda + 1) = 0$$ We want to find the values of $\lambda$ that satisfy this equation. 2. **Formula and rules:** The determinant of a matrix equals zero when the matrix is singular, which leads to the characteristic equation for $\lambda$. The factorization shows the roots directly. 3. **Solve the factorized equation:** Set each factor equal to zero: - $\lambda - 3 = 0 \implies \lambda = 3$ - $2\lambda + 1 = 0 \implies 2\lambda = -1 \implies \lambda = -\frac{1}{2}$ 4. **Interpretation:** The determinant equals zero at $\lambda = 3$ and $\lambda = -\frac{1}{2}$. **Final answer:** $$\boxed{\lambda = 3 \text{ or } \lambda = -\frac{1}{2}}$$ --- The additional information about the circle, triangle, and square is unrelated to the determinant problem and is not addressed here as per instructions.