1. **State the problem:** Find all 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:** The determinant of an upper triangular matrix is the product of its diagonal entries. 3. **Apply the formula:** $$\det = (\omega - 1)(\omega - 2)(\omega - 3)$$ 4. **Set the determinant equal to zero:** $$ (\omega - 1)(\omega - 2)(\omega - 3) = 0 $$ 5. **Solve for $\omega$:** The product is zero if any factor is zero: $$ \omega - 1 = 0 \implies \omega = 1 $$ $$ \omega - 2 = 0 \implies \omega = 2 $$ $$ \omega - 3 = 0 \implies \omega = 3 $$ 6. **Final answer:** $$ \boxed{\omega = 1, 2, 3} $$ These are the values of $\omega$ for which the determinant is zero.