One. (20 points) Let $k, t$ be constants, and the equation about $x$
$$
k x^{2}+(3-3 k) x+2 k-6=0
$$

has only integer roots. The quadratic equation about $y$
$$
(k+3) y^{2}-15 y+t=0
$$

has two positive integer roots $y_{1}, y_{2}$.
(1) Find the value of $k$;
(2) For what value of $t$ is $y_{1}^{2}+y_{2}^{2}$ minimized?
When $k=0$, $x=2$, which satisfies the problem.
Thus, $3 y^{2}-15 y+t=0$.
By the relationship between roots and coefficients, we have
$$
y_{1}+y_{2}=5, y_{1} y_{2}=\frac{t}{3} \text {. }
$$

Then $\left(y_{1}, y_{2}, t\right)=(4,1,2)$ or $(3,2,18)$
$\Rightarrow y_{1}^{2}+y_{2}^{2}=17$ or 13.
When $k \neq 0$, it is easy to get
$$
\begin{array}{l}
(x-2)(k x+3-k)=0 \\
\Rightarrow x_{1}=2, x_{2}=\frac{k-3}{k} .
\end{array}
$$

Thus, $k=\frac{3}{1-x_{2}}, k+3=\frac{6-3 x_{2}}{1-x_{2}}$.
$$
\begin{array}{l}
\text { Also, } y_{1}+y_{2}=\frac{15}{k+3}=\frac{5\left(1-x_{2}\right)}{2-x_{2}} \\
=5+\frac{5}{x_{2}-2} \geqslant 2 \\
\Rightarrow\left(x_{2}-2\right) 15 \\
\Rightarrow x_{2}-2=-5,1,5 \\
\Rightarrow x_{2}=-3,3,7 \\
\Rightarrow k=\frac{3}{4},-\frac{3}{2},-\frac{1}{2} \\
\Rightarrow y_{1}+y_{2}=4,10,6 .
\end{array}
$$

When $y_{1}+y_{2}=4$,
$$
\left(y_{1}, y_{2}\right)=(3,1) \text { or }(2,2) \text {. }
$$

Thus, $y_{1}^{2}+y_{2}^{2}=8$ or 10;
When $y_{1}+y_{2}=6$ or 10,
$$
\begin{array}{c}
y_{1}^{2}+y_{2}^{2} \geqslant\left(6-y_{2}\right)^{2}+y_{2}^{2} \\
=2\left(y_{2}-3\right)^{2}+18 \geqslant 18 .
\end{array}
$$

Therefore, the minimum value of $y_{1}^{2}+y_{2}^{2}$ is 8.
At this time, $y_{1}=y_{2}=2, k=\frac{3}{4}$.
$$
\begin{array}{l}
\text { Also, } y_{1} y_{2}=\frac{t}{k+3} \\
\Rightarrow t=(k+3) y_{1} y_{2}=15 .
\end{array}
$$

In summary, the values of $k$ are $0, \frac{3}{4}, -\frac{3}{2}, -\frac{1}{2}$, and when $t=15$,
$y_{1}^{2}+y_{2}^{2}$ has the minimum value.