Three, (25 points) An oil company plans to build three oil storage facilities in a certain area, located somewhere in the east, west, and south of the city. To save on investment, the company has decided to construct a fuel transfer center in a certain location to ensure adequate fuel supply and convenient transportation to each oil storage facility. As shown in Figure 5(a), the distances between oil storage facilities $A$, $B$, and $C$ are $A B=10 \mathrm{~km}$, $B C=6 \mathrm{~km}$, and $C A=8 \mathrm{~km}$.
Figure 5
After surveying and designing, Designer A says: “When the transfer center is built at point $P$ in Figure 5(b), and the routes $P A$, $P B$, and $P C$ satisfy $\angle A P B=\angle B P C=\angle C P A$, the total length of the transportation routes from the transfer center to the three oil storage facilities is the shortest.” Designer B says: “When the transfer center is built at the point $P$ mentioned by Designer A, the sum of the distances from the transfer center to the three oil storage facilities must be greater than $12 \mathrm{~km}$.”

Do you agree with the views of Designer A and B? If you agree, provide your reasons and explain their rationality; if you disagree, please provide your optimal design solution.
Three, the views proposed by designers A and B are both correct.
For designer A, as shown in

Figure 7, if $\triangle P A C$ is rotated $60^{\circ}$ clockwise around point
$C$ to
$\triangle P^{\prime} A^{\prime} C$, then
$P^{\prime} C=P C$,
$\angle P C P^{\prime}=60^{\circ}$.
Thus, $\triangle P C P$ is an equilateral

triangle, so $\angle C P P^{\prime}=60^{\circ}$.
Also, $\angle A P B=\angle B P C=\angle C P A=120^{\circ}$, hence $B 、 P 、 P$ 、 $A^{\prime}$ are collinear.
At this point, $P A+P B+P C=B P+P P^{\prime}+P A^{\prime}=B A^{\prime}$ is minimized.
Therefore, the conclusion stated by designer A is correct.
For designer B, when the logistics center is built at the point $P$ mentioned by designer A, since $B C=6, C A=8, A B=10$, $\triangle A B C$ is a right triangle, and $\angle C=90^{\circ}$.
Draw $A^{\prime} H \perp B C$, intersecting the extension of $B C$ at point $H$.
Thus, $A^{\prime} B^{2}=A^{\prime} H^{2}+B H^{2}$.
From $\angle A C A^{\prime}=60^{\circ}$, we get
$\angle A^{\prime} C H=30^{\circ}, A^{\prime} H=\frac{1}{2} A^{\prime} C=\frac{1}{2} A C=4$.
In the right triangle $\triangle A^{\prime} C H$,
$$
C H=\sqrt{A^{\prime} C^{2}-A^{\prime} H^{2}}=\sqrt{64-16}=4 \sqrt{3} \text {. }
$$

Then $B H=B C+C H=6+4 \sqrt{3}$.
Thus, $A^{\prime} B^{2}=4^{2}+(6+4 \sqrt{3})^{2}=100+48 \sqrt{3}$, which means
$$
A^{\prime} B=\sqrt{100+48 \sqrt{3}} \text {. }
$$

Since $100+48 \sqrt{3}>144=12^{2}$, then $A^{\prime} B>12$.
Therefore, the view of designer B is also correct.