Three. (25 points) Let $a$ be a positive integer. If the quadratic function $y=2 x^{2}+(2 a+23) x+10-7 a$ and the reciprocal function $y=\frac{11-3 a}{x}$ have a common integer point (a point where both the x-coordinate and y-coordinate are integers), find the value of $a$ and the corresponding common integer point.

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Three, combining the two equations and eliminating $y$ yields
$$
2 x^{2}+(2 a+23) x+10-7 a=\frac{11-3 a}{x} \text {, }
$$

which simplifies to $2 x^{3}+(2 a+23) x^{2}+(10-7 a) x+3 a-11=0$.
Factoring gives
$$
(2 x-1)\left[x^{2}+(a+12) x+11-3 a\right]=0 \text {. }
$$

If the graphs of the two functions have a common integer point, then equation (1) must have an integer root. Therefore, the quadratic equation in $x$
$$
x^{2}+(a+12) x+11-3 a=0
$$

must have an integer root. Thus, the discriminant $\Delta$ of equation (2) must be a perfect square.
$$
\begin{array}{l}
\text { and } \Delta=(a+12)^{2}-4(11-3 a) \\
=a^{2}+36 a+100=(a+18)^{2}-224 \text {, } \\
\end{array}
$$

Therefore, $(a+18)^{2}-224$ must be a perfect square.
Let $(a+18)^{2}-224=k^{2}$ (where $k$ is a non-negative integer), then
$$
(a+18)^{2}-k^{2}=224 \text {, }
$$

which simplifies to $(a+18+k)(a+18-k)=224$.
Following the solution method of the third problem in the A paper, we get
$$
\left\{\begin{array} { l } 
{ a = 3 9 , } \\
{ k = 5 5 }
\end{array} \text { or } \left\{\begin{array}{l}
a=12 \text {, } \\
k=26 .
\end{array}\right.\right.
$$

When $a=39$, equation (2) becomes $x^{2}+51 x-106=0$, whose roots are 2 and -53. It is easy to find that the graphs of the two functions have common integer points $(2,-53)$ and $(-53,2)$.

When $a=12$, equation (2) becomes $x^{2}+24 x-25=0$, whose roots are 1 and -25. It is easy to find that the graphs of the two functions have common integer points $(1,-25)$ and $(-25,1)$.