Let R³ denote the three-dimensional space. Take two points $P=(1,2,3)$ and $Q=(4,2,7)$. Let $\mathrm{dist}<!--\; \; -->(X,Y)$ denote the distance between two points X and Y in R³. Let $S=\{X\in <!--\; \in \; -->{\mathbb{R}}^3:(\mathrm{dist}<!--\; \; -->(X,P){)}^2-<!--\; -\; -->(\mathrm{dist}<!--\; \; -->(X,Q){)}^2=50\}$and $T=\{Y\in <!--\; \in \; -->{\mathbb{R}}^3:(\mathrm{dist}<!--\; \; -->(Y,Q){)}^2-<!--\; -\; -->(\mathrm{dist}<!--\; \; -->(Y,P){)}^2=50\}$.

Then which of the following statements is (are) TRUE?

To solve the problem, let us first find the equations representing the sets S and T in the three-dimensional space ${ℝ}^3$.

Let $X=(x,y,z)$ be a point in ${ℝ}^3$.
The two given points are $P=(1,2,3)$ and $Q=(4,2,7)$.

The squared distance from $X$ to $P$ is:
$(\mathrm{dist}(X,P){)}^2=(x-1{)}^2+(y-2{)}^2+(z-3{)}^2$

The squared distance from $X$ to $Q$ is:
$(\mathrm{dist}(X,Q){)}^2=(x-4{)}^2+(y-2{)}^2+(z-7{)}^2$

The set S is defined by the equation:
$(\mathrm{dist}(X,P){)}^2-(\mathrm{dist}(X,Q){)}^2=50$
Substituting the expressions, we get:
$[(x-1{)}^2+(y-2{)}^2+(z-3{)}^2]-[(x-4{)}^2+(y-2{)}^2+(z-7{)}^2]=50$

Expanding and simplifying:

$(-2x+8x)+(-6z+14z)+(1+9-16-49)=50$
$6x+8z-55=50$
$6x+8z=105$

Thus, the set S represents a plane in ${ℝ}^3$ with the equation:
${\Pi }_1:6x+8z=105$

Similarly, the set T is defined by:
$(\mathrm{dist}(Y,Q){)}^2-(\mathrm{dist}(Y,P){)}^2=50$
This is simply the negative of the equation for S. Therefore, the equation for T is:

$-6x-8z+55=50$
$6x+8z=5$

Thus, the set T represents a plane in ${ℝ}^3$ with the equation:
${\Pi }_2:6x+8z=5$

Since both planes have normal vector $\overrightarrow{n}=(6,0,8)$, they are parallel to each other.
The distance $d$ between these two parallel planes is:
$d=\frac{|105-5|}{\sqrt{6^2+0^2+8^2}}=\frac{100}{10}=10$

Now let us analyze each of the statements:

Statement 1: "There is a triangle whose area is 1 and all of whose vertices are from S."
Since S is a two-dimensional plane in three-dimensional space, we can choose any three non-collinear points in this plane to form a triangle. For example, we can easily construct a triangle in S with base 2 and height 1, which gives an area of 1. Hence, this statement is TRUE.

Statement 2: "There are two distinct points L and M in T such that each point on the line segment LM is also in T."
Since T is a plane, it is a convex set. For any two points L and M in a plane, the entire line segment connecting them also lies within that plane. Thus, this statement is TRUE.

Statement 3: "There are infinitely many rectangles of perimeter 48, two of whose vertices are from S and the other two vertices are from T."
Let a rectangle have two vertices A, B in S and the other two vertices C, D in T.
For this figure to be a rectangle, the sides AD and BC must be perpendicular to both parallel planes S and T.
Thus, the length of these sides must equal the perpendicular distance between the two planes, which is $d=10$.
Let the length of the other two sides (AB and CD, lying entirely in S and T respectively) be $w$.
The perimeter of the rectangle is:
$\text{Perimeter}=2(10)+2w=20+2w$
We are given that the perimeter is 48:
$20+2w=48\Rightarrow 2w=28\Rightarrow w=14$
Since the side length $w=14$ is a positive real number, such a rectangle exists.
Moreover, we can rotate and translate this rectangle in any direction parallel to the planes S and T, yielding infinitely many such rectangles. Hence, this statement is TRUE.

Statement 4: "There is a square of perimeter 48, two of whose vertices are from S and the other two vertices are from T."
If the rectangle is a square, then all its sides must be equal.
Therefore, the side perpendicular to the planes must equal the side parallel to the planes, meaning $w=10$.
This would result in a perimeter of:
$\text{Perimeter}=4\times 10=40$
However, the statement requires a perimeter of 48, which would force the side length to be $48/4=12$. But the distance between the planes is fixed at 10, so a square with vertices on S and T must have a side length of exactly 10. Thus, a square of perimeter 48 cannot exist. Hence, this statement is FALSE.