The number $(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}$ is equal to:
(a) $-\sqrt{3}$;
(b) $-\sqrt{2}$;
(c) -2 ;
(d) 1 ;
(e) 2.
The correct option is (c).

Observe that, denoting by $A$ the given expression, we have

$$
\begin{aligned}
A^{2} & =[(\sqrt{6}+\sqrt{2})(\sqrt{3}-2) \sqrt{\sqrt{3}+2}]^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{\sqrt{3}+2})^{2} \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)^{2}(\sqrt{3}+2) \\
& =(\sqrt{6}+\sqrt{2})^{2}(\sqrt{3}-2)[(\sqrt{3}-2)(\sqrt{3}+2)] \\
& =(6+2 \sqrt{12}+2)(\sqrt{3}-2)\left((\sqrt{3})^{2}-2^{2}\right) \\
& =(6+2 \sqrt{12}+2)(\sqrt{3}-2)(-1) \\
& =(8+4 \sqrt{3})(2-\sqrt{3})=4(2+\sqrt{3})(2-\sqrt{3}) \\
& =4\left(2^{2}-(\sqrt{3})^{2}\right)=4 \times 1=4
\end{aligned}
$$

Thus, $A^{2}=4$ and, therefore, $A$ can be 2 or -2. Since $\sqrt{3}-2$ is negative and the other two factors of $A$ are positive, we have that $A$ must be negative, that is, $A=-2$.