10. The sequence of real numbers $a_{0}, a_{1}, a_{2}, \cdots, a_{n}, \cdots$ satisfies the following equation: $a_{0}=a$, where $a$ is a real number,
$$
a_{n}=\frac{a_{n-1} \sqrt{3}+1}{\sqrt{3}-a_{n-1}}, n \in \mathbf{N}
$$

Then $a_{1994}=$
10. $\frac{a+\sqrt{3}}{1-a \sqrt{3}}$.

From the given, we have: $a_{1}=\frac{a \sqrt{3}+1}{\sqrt{3}-a}$,
$$
\begin{array}{l}
a_{2}=\frac{a_{1} \sqrt{3}+1}{\sqrt{3}-a_{1}}=\frac{(a \sqrt{3}+1) \sqrt{3}+(\sqrt{3}-a)}{\sqrt{3}(\sqrt{3}-a)-(a \sqrt{3}+1)}=\frac{a+\sqrt{3}}{1-a \sqrt{3}}, \\
a_{3}=\frac{a_{2} \sqrt{3}+1}{\sqrt{3}-a_{2}}=\frac{(a+\sqrt{3}) \sqrt{3}+(1-a \sqrt{3})}{\sqrt{3}(1-a \sqrt{3})-(a+\sqrt{3})}=-\frac{1}{a} .
\end{array}
$$

From this, we get $a_{6}=-\frac{1}{-\frac{1}{a}}=a$. Therefore, the sequence $a_{0}, a_{1}, a_{2}, \cdots$ is a periodic sequence with a period of 6.
Since $1994=6 \times 332+2$, we have $a_{1994}=a_{2}=\frac{a+\sqrt{3}}{1-a \sqrt{3}}$. Hence, the answer should be $\frac{a+\sqrt{3}}{1-a \sqrt{3}}$.