1. **State the problem:** We want to find the values of $\omega$ such that the determinant of the matrix $$\begin{bmatrix} \omega - 1 & -1 & -2 \\ 0 & \omega - 2 & 2 \\ 0 & 0 & \omega - 3 \end{bmatrix}$$ is zero. 2. **Recall the formula:** For an upper triangular matrix, the determinant is the product of the diagonal entries. 3. **Write the determinant:** $$\det = (\omega - 1)(\omega - 2)(\omega - 3)$$ 4. **Set the determinant equal to zero:** $$ (\omega - 1)(\omega - 2)(\omega - 3) = 0 $$ 5. **Solve for $\omega$:** The product of factors is zero if at least one factor is zero: $$\omega - 1 = 0 \implies \omega = 1$$ $$\omega - 2 = 0 \implies \omega = 2$$ $$\omega - 3 = 0 \implies \omega = 3$$ 6. **Final answer:** $$\boxed{\omega = 1, 2, 3}$$ This means the determinant is zero exactly when $\omega$ is 1, 2, or 3.