Let $$\omega$$ be a complex number such that $$2\omega + 1 = z$$ where $$z = \sqrt{-3}$$. If
$$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -\omega^{2}-1 & \omega^{2} \\ 1 & \omega^{2} & \omega^{7} \end{vmatrix} = 3k$$,
then $$k$$ can be equal to:

We have the relation $$2\omega + 1 = z$$ with $$z = \sqrt{-3}$$. Because $$\sqrt{-3} = \pm i\sqrt3$$, let us first isolate $$\omega$$:

$$2\omega = z-1 \;\;\Longrightarrow\;\; \omega = \dfrac{z-1}{2}.$$

If we choose the principal value $$z = i\sqrt3,$$ then

$$\omega = \dfrac{i\sqrt3-1}{2} = -\dfrac12 + i\dfrac{\sqrt3}{2}.$$

This number is well known: it is a primitive cube-root of unity. Hence the standard identities hold:

$$\omega^3 = 1,\qquad \omega \neq 1,\qquad 1+\omega+\omega^2 = 0.$$

From $$1+\omega+\omega^2 = 0$$ we also get

$$\omega^2 = -1-\omega.$$

Because $$\omega^3 = 1$$, every higher power may be reduced modulo 3; in particular

$$\omega^6 = 1,\qquad\omega^7 = \omega^{6}\omega = \omega.$$

Now we evaluate the determinant

$$ \begin{vmatrix} 1 & 1 & 1\\[2pt] 1 & -\omega^{2}-1 & \omega^{2}\\[2pt] 1 & \omega^{2} & \omega^{7} \end{vmatrix}. $$

Substituting $$\omega^{7}=\omega$$ gives

$$ \begin{vmatrix} 1 & 1 & 1\\[2pt] 1 & -(\omega^{2}+1) & \omega^{2}\\[2pt] 1 & \omega^{2} & \omega \end{vmatrix}. $$

We expand along the first row using the formula

$$\begin{vmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{vmatrix} =a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31}) +a_{13}(a_{21}a_{32}-a_{22}a_{31}).$$

Here $$a_{11}=a_{12}=a_{13}=1,$$ so we compute the three minors one by one:

First minor:

$$ a_{22}a_{33}-a_{23}a_{32} =\bigl(-(\omega^{2}+1)\bigr)\omega-\omega^{2}\cdot\omega^{2} =-\omega(\omega^{2}+1)-\omega^{4}. $$

Since $$\omega^{4}=\omega^{3}\omega=1\cdot\omega=\omega,$$ this becomes

$$-\omega^{3}-\omega-\omega=-1-2\omega.$$

Second minor:

$$ a_{21}a_{33}-a_{23}a_{31} =1\cdot\omega-\omega^{2}\cdot1=\omega-\omega^{2}. $$

Third minor:

$$ a_{21}a_{32}-a_{22}a_{31} =1\cdot\omega^{2}-\bigl(-(\omega^{2}+1)\bigr)\cdot1 =\omega^{2}+\omega^{2}+1=2\omega^{2}+1. $$

Putting these pieces into the expansion gives

$$ \begin{aligned} \det &= 1\bigl(-1-2\omega\bigr)-1\bigl(\omega-\omega^{2}\bigr)+1\bigl(2\omega^{2}+1\bigr)\\[4pt] &=(-1-2\omega)-\omega+\omega^{2}+2\omega^{2}+1\\[4pt] &=-1-2\omega-\omega+\omega^{2}+2\omega^{2}+1\\[4pt] &=-3\omega+3\omega^{2}\\[4pt] &=3\bigl(\omega^{2}-\omega\bigr). \end{aligned} $$

The problem states that this determinant equals $$3k,$$ so

$$3k = 3(\omega^{2}-\omega)\;\;\Longrightarrow\;\; k = \omega^{2}-\omega.$$

Using $$\omega^{2} = -1-\omega$$ we write

$$k = (-1-\omega)-\omega = -1-2\omega.$$

But from the very first relation $$2\omega+1 = z,$$ multiplying both sides by $$-1$$ gives

$$-2\omega-1 = -z.$$

Hence

$$k = -1-2\omega = -z.$$

Among the given choices the expression $$-z$$ appears in Option A.

Hence, the correct answer is Option A.