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

\(\begin{aligned}& \mathrm{S}=\left\{\mathrm{X}:(\mathrm{XP})^2-(\mathrm{XQ})^2=50\right\} \\& \mathrm{T}=\left\{\mathrm{Y}:(\mathrm{YQ})^2-(\mathrm{YP})^2=50\right\}\end{aligned}\)
for finding \(S \equiv X(x, y, z)\) and for \(T \equiv Y(x, y, z)\)
\(\begin{aligned}&\left((x-1)^2+(y-1)^2+(\mathrm{z}-1)^2\right)-\left((\mathrm{x}-4)^2+(\mathrm{y}-2)^2+(\mathrm{z}-7)^2\right)=50 \\& \Rightarrow \quad \mathrm{S}=\{(\mathrm{x}, \mathrm{y}, \mathrm{z}): 6 \mathrm{x}+8 \mathrm{z}=105\} \\& \mathrm{T}=\{(\mathrm{x}, \mathrm{y}, \mathrm{z}): 6 \mathrm{x}+8 \mathrm{z}=5\}\end{aligned}\)
Since \(\mathrm{S}\) and \(\mathrm{T}\) both are plane ;
(1) There exist a triangle in plane \(\mathrm{S}\) whose area \(=1\) (always)
(2) \(\mathrm{L} ~\&~ \mathrm{M}\) lies on plane \(\mathrm{T}\), hence line segment joining \(\mathrm{L} \& \mathrm{M}\) will lie on plane \(\mathrm{T}\).
(3) Distance between \(\mathrm{S} \& \mathrm{~T}\)
\(\mathrm{d}=\left|\frac{105-5}{10}\right|=10\)
Hence for rectangle of perimeter 48 can exist.
(4) For Square

There will be infinite such rectangle possible.
Hence Answers 1,2,3,4 are correct.