$\mathrm{}\mathrm{If}A=\left[\begin{array}{ccc}0& 0& \sqrt{3}\\ 0& \sqrt{3}& 0\\ \sqrt{3}& 0& 0\end{array}\right],\mathrm{then}-\mathrm{adj}A\mathrm{is}\mathrm{equal}\mathrm{to}$

The correct answer is 27.

Step-by-step Explanation:
First, let us clarify the notation. The expression printed as $-\mathrm{adj}A$ represents the determinant of the adjoint of matrix $A$, which is denoted by $\left|\mathrm{adj}A\right|$.

The given matrix $A$ is:
$$A=\left[\begin{array}{ccc}0& 0& \sqrt{3}\\ 0& \sqrt{3}& 0\\ \sqrt{3}& 0& 0\end{array}\right]$$

Step 1: Find the determinant of matrix $A$ (denoted as $\left|A\right|$)
We expand the determinant along the first row:
$$\left|A\right|=0\cdot \left|\begin{array}{cc}\sqrt{3}& 0\\ 0& 0\end{array}\right|-0\cdot \left|\begin{array}{cc}0& 0\\ \sqrt{3}& 0\end{array}\right|+\sqrt{3}\cdot \left|\begin{array}{cc}0& \sqrt{3}\\ \sqrt{3}& 0\end{array}\right|$$

Simplifying the terms:
$$\left|A\right|=0-0+\sqrt{3}(0-(\sqrt{3}\cdot \sqrt{3}\left)\right)$$
$$\left|A\right|=\sqrt{3}(-3)=-3\sqrt{3}$$

Step 2: Apply the property of the determinant of an adjoint matrix
For any square matrix $A$ of order $n$, the determinant of its adjoint is given by:
$$\left|\mathrm{adj}A\right|={\left|A\right|}^{n-1}$$

Since $A$ is a $3\times 3$ matrix, its order is $n=3$.
Therefore:
$$\left|\mathrm{adj}A\right|={\left|A\right|}^{3-1}={\left|A\right|}^2$$

Step 3: Calculate the final value
Substitute the value of $\left|A\right|$ into the equation:
$$\left|\mathrm{adj}A\right|={(-3\sqrt{3})}^2$$
$$\left|\mathrm{adj}A\right|={(-3)}^2\cdot {\left(\sqrt{3}\right)}^2=9\cdot 3=27$$

Thus, the value of the determinant is 27.