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 $\mathbb{R}^{3}$. Let $S = \{X \in \mathbb{R}^{3} : (\text{dist}(X, P))^{2} - (\text{dist}(X, Q))^{2} = 50\}$ 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?

$P=(1,2,3), \quad Q=(4,2,7)$

$S = \left\{X : (\operatorname{dist}(X,P))^2 - (\operatorname{dist}(X,Q))^2 = 50\right\}$

$T = \left\{Y : (\operatorname{dist}(Y,Q))^2 - (\operatorname{dist}(Y,P))^2 = 50\right\}$

$\text{To find: which of } (1),\ (2),\ (3),\ (4) \text{ are TRUE?}$

$|X-P|^2 - |X-Q|^2 = 2X\cdot(Q-P) + |P|^2 - |Q|^2$

$Q-P = (3,0,4), \quad |P|^2 = 14, \quad |Q|^2 = 69$

$|X-P|^2 - |X-Q|^2 = 6x + 8z - 55$

$S: 6x + 8z - 55 = 50 \Rightarrow 6x + 8z = 105$

$T: 6x + 8z - 55 = -50 \Rightarrow 6x + 8z = 5$

$\vec{n} = (6,0,8) \parallel (3,0,4) = \overrightarrow{PQ}$

$\Rightarrow S \parallel T \quad (\text{both} \perp \overrightarrow{PQ})$

$d = \dfrac{|105 - 5|}{\sqrt{6^2 + 8^2}} = \dfrac{100}{10} = 10$

$S,\ T \text{ are parallel planes, distance } 10 \text{ apart}$

$S \text{ is a plane} \Rightarrow \text{triangles of every area} \Rightarrow (1)\ \checkmark$

$T \text{ is a plane (convex)} \Rightarrow LM \subset T \Rightarrow (2)\ \checkmark$

$\text{Connecting side } \perp \text{planes}: \text{length } 10$

$\text{In-plane sides } 14 \Rightarrow 2(10 + 14) = 48$

$\text{Slide the in-plane side along } S \Rightarrow \text{infinitely many} \Rightarrow (3)\ \checkmark$

$\text{Square side } 12; \text{ connecting side } = 10\hat{n} + 2\sqrt{11}\,\hat{f}$

$\text{In-plane side } 12\hat{e},\ \hat{e}\perp\hat{f} \Rightarrow \text{square exists}$

$\text{Two vertices on } S, \text{ two on } T \Rightarrow (4)\ \checkmark$

$\boxed{(1),\ (2),\ (3),\ (4) \text{ all TRUE}}$