Three. (This question is worth 25 points) $a$ is a real number greater than zero. It is known that there exists a unique real number $k$ such that the quadratic equation $x^{2}+\left(k^{2}+a k\right) x+1999+k^{2}+a k=0$ has two roots that are both prime numbers. Find the value of $a$.
Three, let the two prime roots of the equation be \( p \) and \( q \). By the relationship between the roots and coefficients of a quadratic equation, we have
\[
\begin{array}{l}
p+q=-k^{2}-a k, \\
p q=1999+k^{2}+a k . \\
(1)+(2), \text { we get } p+q+p q=1999 . \\
\therefore(p+1)(q+1)=2^{4} \times 5^{3} .
\end{array}
\]

From (3), it is clear that \( p \) and \( q \) cannot be 2, so they must be odd.
\(\therefore \frac{p+1}{2}\) and \(\frac{q+1}{2}\) are both integers, and
\[
\frac{p+1}{2} \cdot \frac{q+1}{2}=2^{2} \times 5^{3} \text {. }
\]

If \(\frac{p+1}{2}\) is odd, it must be \(5^{r} (r=1,2,3)\). Then \(p=2 \times 5^{r}-1\) is a composite number, which is a contradiction.

Similarly, \(\frac{q+1}{2}\) must also be even. Therefore, \(\frac{p+1}{4}\) and \(\frac{q+1}{4}\) are both integers, and
\[
(p+1)(q+1)=5
\]

Assume \(p=q\), then \(\frac{p+1}{4}=1\) or 5.
If \(\frac{p+1}{4}=1\), then \(\frac{q+1}{4}=5^{-1}\), giving \(p=3, q=499\), which are prime numbers.
\[
\text { If } \frac{p+1}{4}=5 \text {, then } \frac{q+1}{4}=5^{2} \text {, giving } p=19, q=99 \text {, which are }
\]

composite numbers, which does not meet the problem's requirements.
In summary, \(p=3, q=499\).
Substituting into (1), we get
\[
k^{2}+a k+502=0 \text {. }
\]

According to the problem, equation (4) has a unique real solution.
\[
\therefore \Delta=a^{2}-4 \times 502=0 \text {. }
\]

Thus, \(a=2 \sqrt{502}\).
(Provided by Chen Chuanli)