1. Interpret the given condition

   We are given $2\omega + 1 = z, \quad z = \sqrt{-3} = i\sqrt{3}.$

   Hence $2\omega = z-1 \implies \omega = \frac{z-1}{2}.$

   Since $z=i\sqrt{3}$, this is $\omega = \frac{-1+i\sqrt{3}}{2}.$

   This is a cube root of unity: $\omega^3=1, \quad \omega\ne 1.$ Therefore, $1+\omega+\omega^2=0, \quad \omega^2+\omega=-1.$

2. Reduce higher powers

   Because $\omega^3=1$, $\omega^7=\omega^{6+1}=\omega.$

   So the determinant becomes

   1 & 1 & 1 \\ 1 & -\omega^2-1 & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}.$$

3. Use the relation $-\omega^2-1=\omega$

   From $1+\omega+\omega^2=0 \implies -\omega^2-1=\omega.$

   Thus

   1 & 1 & 1 \\ 1 & \omega & \omega^2 \\ 1 & \omega^2 & \omega \end{vmatrix}.$$

4. Evaluate the determinant

   Apply row operations: $R_2 \to R_2-R_1, \quad R_3 \to R_3-R_1.$

   Then

   1 & 1 & 1 \\ 0 & \omega-1 & \omega^2-1 \\ 0 & \omega^2-1 & \omega-1 \end{vmatrix}.$$ Expanding along the first column, $$D=\begin{vmatrix} \omega-1 & \omega^2-1 \\ \omega^2-1 & \omega-1 \end{vmatrix}.$$ So $$D=(\omega-1)^2-(\omega^2-1)^2.$$ Factor as difference of squares: $$D=\big[(\omega-1)-(\omega^2-1)\big]\big[(\omega-1)+(\omega^2-1)\big].$$ That is, $$D=(\omega-\omega^2)(\omega+\omega^2-2).$$ Using $$\omega+\omega^2=-1,$$ we get $$D=(\omega-\omega^2)(-3).$$ Hence $$D=-3(\omega-\omega^2)=3(\omega^2-\omega).$$

5. Find $\omega^2-\omega$

   For cube roots of unity, $\omega=\frac{-1+i\sqrt{3}}{2}, \quad \omega^2=\frac{-1-i\sqrt{3}}{2}.$

   Therefore, $\omega^2-\omega=\frac{-1-i\sqrt{3}}{2}-\frac{-1+i\sqrt{3}}{2}=-i\sqrt{3}.$

   Since $z=i\sqrt{3},$ we have $\omega^2-\omega=-z.$

   Therefore, $D=3(-z).$

   Comparing with $D=3k,$ we get $k=-z.$

6. Match with the options

   $k=-z$ corresponds to Option D.

Comparison with stored answer: Stored correct answer is D, which matches our result.