Three. (50 points) Given $a, b, c, d \in \mathbf{R}_{+}, k_{1}, k_{2}, k_{3}, k_{4} \in \mathbf{N}_{+}$, and
$$
\begin{array}{l}
b+c+d=k_{1} a, c+d+a=k_{2} b, \\
d+a+b=k_{3} c, a+b+c=k_{4} d .
\end{array}
$$

Try to find the set of values for $k_{1}+k_{2}+k_{3}+k_{4}$.
Three, note that
$$
\begin{array}{l}
k_{1}+k_{2}+k_{3}+k_{4} \\
=\frac{b+c+d}{a}+\frac{c+d+a}{b}+\frac{d+a+b}{c}+\frac{a+b+c}{d} .
\end{array}
$$

Since the above expression is symmetric in $a, b, c, d$, we can assume without loss of generality that $a \geqslant b \geqslant c \geqslant d$. Then $k_{1}=\frac{b+c+d}{a} \leqslant 3$.
(1) If $k_{1}=3$, then $a=b=c=d$, at this time,
$$
k_{1}+k_{2}+k_{3}+k_{4}=12 \text {. }
$$
(2) If $k_{1}=2$, i.e.,
$$
b+c+d=2 a \text {. }
$$

Then $k_{2}=\frac{c+d+a}{b}=\frac{\frac{3}{2}(c+d)+\frac{1}{2} b}{b}$.
So, $1<k_{2} \leqslant \frac{7}{2}$.
(i) If $k_{2}=3$, i.e.,
$$
c+d+a=3 b \text {. }
$$

From equations (1) and (2), we get $3 a=4 b, c+d=\frac{5}{3} b$. At this time,
$k_{3}=\frac{a+b+d}{c}=\frac{4 b}{c}-1$,
$k_{4}=\frac{a+b+c}{d}=\frac{4 b}{d}-1$.
Let $\frac{4 b}{d}=m, \frac{4 b}{c}=n$. Then $m, n \in \mathbf{N}_{+}, m \geqslant n \geqslant 4$.
Also, from $c+d=\frac{5}{3} b$ we get $\frac{4 b}{m}+\frac{4 b}{n}=\frac{5}{3} b$.
Simplifying, we get $(5 m-12)(5 n-12)=12^{2}$.
Since $5 m-12 \geqslant 5 n-12 \geqslant 8$, we have
$\left\{\begin{array}{l}5 m-12=18, \\ 5 n-12=8 .\end{array}\right.$ Solving, we get $\left\{\begin{array}{l}m=6, \\ n=4 .\end{array}\right.$ Thus,
$$
k_{1}+k_{2}+k_{3}+k_{4}=2+3+(4-1)+(6-1)=13 \text {. }
$$
(ii) If $k_{2}=2$, i.e.,
$c+d+a=2 b$.
From equations (1) and (3), we get $a=b, c+d=b$. At this time,
$$
k_{3}=\frac{a+b+d}{c}=\frac{3 b}{c}-1, k_{4}=\frac{3 b}{d}-1 \text {. }
$$

Similarly, we solve to get $k_{1}+k_{2}+k_{3}+k_{4}=18$ or 14.
(3) If $k_{1}=1$, i.e.,
$$
b+c+d=a \text {. }
$$

Then $k_{2}=\frac{c+d+a}{b}=1+\frac{2(c+d)}{b}$.
So, $1<k_{2} \leqslant 5$.
(i) If $k_{2}=5$, at this time, $b=c=d, a=3 b$, we get $k_{3}=$ $k_{4}=5$.
Thus, $k_{1}+k_{2}+k_{3}+k_{4}=1+5+5+5=16$.
(ii) If $k_{2}=4$, i.e.,
$$
c+d+a=4 b \text {. }
$$

From equations (4) and (5), we get $a=\frac{5}{2} b, c+d=\frac{3}{2} b$. At this time,
$$
k_{3}=\frac{5 b}{c}-1, k_{4}=\frac{5 b}{d}-1 \text {. }
$$

Similarly, we solve to get $k_{1}+k_{2}+k_{3}+k_{4}=18$.
(iii) If $k_{2}=3$, i.e.,
$$
c+d+a=3 b \text {. }
$$

From equations (4) and (6), we get $a=2 b, c+d=b$. At this time,
$$
k_{3}=\frac{4 b}{c}-1, k_{4}=\frac{4 b}{d}-1 \text {. }
$$

Similarly, we solve to get $k_{1}+k_{2}+k_{3}+k_{4}=27$ or 20 or 18.
(iv) If $k_{2}=2$, i.e.,
$$
c+d+a=2 b \text {. }
$$

From equations (4) and (7), we get $2 a=3 b, c+d=\frac{1}{2} b$. At this time,
$$
k_{3}=\frac{3 b}{c}-1, k_{4}=\frac{3 b}{d}-1 \text {. }
$$

Similarly, we solve to get
$k_{1}+k_{2}+k_{3}+k_{4}=50$ or 33 or 28 or 26 or 25.
In summary, the set of values is
$\{12,13,14,16,18,20,25,26,27,28,33,50\}$.