Let $\mathbb{R}$ denote the three-dimensional space. Take two points $P = (1, 2, 3)$ and $Q = (4, 2, 7)$. Let $\text{dist}\,(X, Y)$ denote the distance between two points $X$ and $Y$ in $\mathbb{R}^{3}$. $S = \{X \in \mathbb{R}^{3} : (\text{dist}\,(X, P))^{2} - (\text{dist}\,(X, Q))^{2} = 50\}$ and $T = \{Y \in \mathbb{R}^{3} : (\text{dist}\,(Y, Q))^{2} - (\text{dist}\,(Y, P))^{2} = 50\}$. Then which of the following statements is (are) TRUE?

$P = (1,2,3), \qquad Q = (4,2,7)$ $S: (\operatorname{dist}(X,P))^2 - (\operatorname{dist}(X,Q))^2 = 50$ $T: (\operatorname{dist}(Y,Q))^2 - (\operatorname{dist}(Y,P))^2 = 50$ $\text{Which of the statements (1)-(4) is/are TRUE?}$ $1.\ \text{expand } S, T$ $2.\ \text{find the distance between them}$ $3.\ \text{judge } (1)\text{-}(4)$ $|X-P|^2 - |X-Q|^2 = 2X\cdot(Q-P) + |P|^2 - |Q|^2$ $= 2(3x+4z) + 14 - 69 = 50 \Rightarrow 6x+8z = 105$ $|Y-Q|^2 - |Y-P|^2 = -2(3x+4z) + (69-14) = 50$ $\Rightarrow 6x+8z = 5$ $S: 6x+8z=105, \quad T: 6x+8z=5 \quad (\text{parallel, normal } (6,0,8))$ $d(S,T) = \dfrac{|105-5|}{\sqrt{6^2+8^2}} = \dfrac{100}{10} = 10$ $\vec{PQ} = (3,0,4) \parallel (6,0,8), \qquad |PQ| = 5$ $(1):\ S \text{ is a whole plane} \Rightarrow \text{a triangle of area } 1 \text{ lies in } S \ \text{✓}$ $(2):\ T \text{ is a plane (convex)} \Rightarrow \text{every point of } LM \text{ is in } T \ \text{✓}$ $(3):\ \text{perimeter } 48 \Rightarrow \text{adjacent sides sum to } 24$ $\text{join side} \ge d = 10; \ 10 + 14 = 24 \Rightarrow 14\times 10 \text{ rectangle}$ $\text{slide/rotate in } S \Rightarrow \text{infinitely many} \ \text{✓}$ $(4):\ \text{perimeter } 48 \Rightarrow \text{side } 12$ $\text{join side } 12: \ \text{normal component } 10, \ \text{in-plane } \sqrt{12^2-10^2} = \sqrt{44}$ $\Rightarrow \text{a } 12\times 12 \text{ square exists} \ \text{✓}$ $\boxed{\text{Statements } (1),\ (2),\ (3),\ (4) \text{ are all TRUE}}$