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

Write $$X=(x,y,z)$$. Using the distance formula,

$$(dist(X,P))^{2}- (dist(X,Q))^{2}$$
$$=(x-1)^{2}+(y-2)^{2}+(z-3)^{2}-\left[(x-4)^{2}+(y-2)^{2}+(z-7)^{2}\right]$$
$$=\left[(x-1)^{2}-(x-4)^{2}\right]+\left[(z-3)^{2}-(z-7)^{2}\right]$$
$$=(6x-15)+(8z-40)=6x+8z-55.$$ Therefore

Set $$S$$ : $$6x+8z-55=50\;\Longrightarrow\;6x+8z=105$$.

Set $$T$$ : $$-(6x+8z-55)=50\;\Longrightarrow\;6x+8z=5$$.

Hence $$S$$ and $$T$$ are two parallel planes perpendicular to the vector $$(6,0,8)$$. Their separation is

$$\frac{|105-5|}{\sqrt{6^{2}+0^{2}+8^{2}}}=\frac{100}{10}=10.$$

Checking the options

Option A  In the infinite plane $$6x+8z=105$$ we can choose any three non-collinear points. Scale them so that the area becomes exactly $$1$$ (e.g. first pick a triangle of area $$k$$ and then take all its vertices closer to their centroid by a factor $$\sqrt{k}$$). Thus a triangle of area $$1$$ with all vertices in $$S$$ exists. Option A is true.

Option B  Because $$T$$ is a plane, pick any two distinct points $$L,M\in T$$; the entire segment $$LM$$ lies in the same plane and therefore in $$T$$. Option B is true.

Option C  Let $$\mathbf n=(6,0,8)$$ and $$\|\mathbf n\|=10$$. For any point $$A\in S$$ define $$D=A-\mathbf n$$ (so $$D\in T$$). Choose a vector $$\mathbf v$$ lying in the plane (orthogonal to $$\mathbf n$$) with length $$14$$. Put $$B=A+\mathbf v$$ and $$C=D+\mathbf v$$. Then $$AB=CD=14$$ lie in $$S$$, $$BC=DA=10$$ join the two planes, and the quadrilateral $$ABCD$$ is a rectangle of perimeter $$2(14+10)=48$$ having two vertices in $$S$$ and two in $$T$$. Because we may pick $$A$$ anywhere in $$S$$ and rotate $$\mathbf v$$ freely in the plane, infinitely many such rectangles exist. Option C is true.

Option D  A square of perimeter $$48$$ has side $$12$$. If two consecutive vertices are in different planes, every side must join the planes, so each side’s normal component equals the plane distance $$10$$. Hence the in-plane component of a side has length $$\sqrt{12^{2}-10^{2}}=2\sqrt{11}$$. Take two adjacent sides $$\mathbf a,\mathbf b$$; their normal components are $$\pm10$$ along the same normal, while their in-plane parts are orthogonal. The dot product condition for a square demands $$\mathbf a\cdot\mathbf b=0$$, that is
$$-10^{2} + (\text{dot product of in-plane parts}) = 0 \;\Longrightarrow\; (\text{dot product of in-plane parts}) = 100.$$
But the two in-plane vectors each have magnitude $$2\sqrt{11}$$, so their dot product is at most $$(2\sqrt{11})^{2}=44\lt100.$$ Therefore such a square cannot be formed. Option D is false.

The true statements are: Option A, Option B, and Option C.

Option A, Option B, Option C