One, (20 points) Given the quadratic function $y=x^{2}-2 k x+$ $3 k+1$, the independent variable $x$ and real numbers $a, b$ satisfy $4 a^{2}+9 b^{2}=$ $x+6 a b=2, y_{\text {minimum }}=1$: Find the value of the constant $k$.
Given:
\[4 a^{2}-6 a b+9 b^{2}=x\].
Also, 
\[4 a^{2}+9 b^{2}=2\],
Subtracting (1) from (2) gives:
\[2 a \cdot 3 b=2-x\].
From equations (2) and (3), we get:
\[
\begin{array}{l}
(2 a+3 b)^{2}=4 a^{2}+9 b^{2}+12 a b \\
=2+2(2-x)=6-2 x .
\end{array}
\]

Thus, 
\[2 a+3 b= \pm \sqrt{6-2 x}\].
Therefore, \(2 a\) and \(3 b\) are the two real roots of the quadratic equation in \(t\):
\[t^{2} \mp \sqrt{6-2 x} t+2-x=0\].
Hence, 
\[\Delta=(\mp \sqrt{6-2 x})^{2}-4(2-x)=2 x-2 \geqslant 0\].
Thus, 
\[x \geqslant 1\].
Also, 
\[6-2 x \geqslant 0\], so 
\[x \leqslant 3\].
Therefore, the range of \(x\) in the quadratic function \(y=x^{2}-2 k x+3 k+1\) is 
\[1 \leqslant x \leqslant 3\].

The vertex of the quadratic function \(y=x^{2}-2 k x+3 k+1\) is \(\left(k,-k^{2}+3 k+1\right)\).
(1) If \(k < 1\), the minimum value of \(y\) occurs at \(x=1\).

Thus, 
\[y_{\text {min }}=1^{2}-2 k \cdot 1+3 k+1=1\].
Solving for \(k\), we get \(k=-1\).

(2) If \(1 \leqslant k \leqslant 3\), the minimum value of \(y\) occurs at \(x=k\).

Thus, 
\[y_{\text {min }}=k^{2}-2 k \cdot k+3 k+1=1\].
Solving for \(k\), we get \(k=1\).

(3) If \(k > 3\), the minimum value of \(y\) occurs at \(x=3\).

Thus, 
\[y_{\text {min }}=3^{2}-2 k \cdot 3+3 k+1=1\].
Solving for \(k\), we get \(k=3\) (discard).

In conclusion, the constant \(k\) can be \(-1\) and \(3\).