Parallel Planes Defined by Difference of Squared Distances

To determine which of the given statements are true, we first determine the geometric representations of the sets $S$ and $T$.

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

Step 1: Equations of Sets $S$ and $T$

For set $S$: $(\text{dist}(X, P))^2 - (\text{dist}(X, Q))^2 = 50$

Expanding the distance terms: $\left[(x - 1)^2 + (y - 2)^2 + (z - 3)^2\right] - \left[(x - 4)^2 + (y - 2)^2 + (z - 7)^2\right] = 50$

Simplifying: $\left[(x - 1)^2 - (x - 4)^2\right] + \left[(z - 3)^2 - (z - 7)^2\right] = 50$ $(6x - 15) + (8z - 40) = 50$ $6x + 8z - 55 = 50 \implies 6x + 8z = 105$

Thus, $S$ is a plane given by the equation: $S : 3x + 4z = \frac{105}{2}$

For set $T$: $(\text{dist}(Y, Q))^2 - (\text{dist}(Y, P))^2 = 50$

This is equivalent to: $- \left[(\text{dist}(Y, P))^2 - (\text{dist}(Y, Q))^2\right] = 50$ $-(6x + 8z - 55) = 50 \implies 6x + 8z = 5$

Thus, $T$ is a plane given by the equation: $T : 3x + 4z = \frac{5}{2}$

Step 2: Relationship Between Planes $S$ and $T$

Both $S$ and $T$ are parallel planes with a common normal vector $\vec{n} = (3, 0, 4)$.

The perpendicular distance $d$ between the parallel planes $S$ and $T$ is: $d = \frac{\left|\frac{105}{2} - \frac{5}{2}\right|}{\sqrt{3^2 + 0^2 + 4^2}} = \frac{50}{5} = 10$

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Step 3: Analysis of the Options

• Option A: Since $S$ is a two-dimensional plane in $\mathbb{R}^3$, we can choose any three non-collinear points in $S$ to form a triangle of any arbitrary positive area, including area $1$. Statement A is TRUE.

• Option B: Since $T$ is a plane, it is a convex set. For any two distinct points $L, M \in T$, the entire line segment $LM$ lies completely within the plane $T$. Statement B is TRUE.

• Option C: Consider a rectangle $ABCD$ where $A, B \in S$ and $C, D \in T$. Let the length of side $AD$ (connecting a point on plane $S$ to a point on plane $T$) be $h$. Since the minimum distance between the two planes is $10$, we must have $h \ge 10$. For $ABCD$ to have a perimeter of $48$: $2(AB + AD) = 48 \implies AB + h = 24 \implies AB = 24 - h$ Since $AB > 0$, any $h \in [10, 24)$ yields a valid rectangle length $AB = 24 - h > 0$. Since there are infinitely many choices for $h$ in this interval, as well as infinitely many choices for the location and orientation of point $A \in S$, there exist infinitely many such rectangles. Statement C is TRUE.

• Option D: For a square of perimeter $48$, each side length must be $\frac{48}{4} = 12$. Choosing $AD = 12$ and $AB = 12$: Since $12 \ge 10$ (the perpendicular distance between planes $S$ and $T$), we can orient a vector $\vec{AD}$ of length $12$ connecting $A \in S$ to $D \in T$ such that its angle $\theta$ with the normal $\hat{n}$ satisfies $\cos\theta = \frac{10}{12} = \frac{5}{6}$. We can then choose a vector $\vec{AB}$ in the plane $S$ of length $12$ that is perpendicular to $\vec{AD}$. Completing the rectangle with $C = B + \vec{AD} \in T$ forms a square of side $12$ and perimeter $48$. Statement D is TRUE.

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Conclusion

All four statements A, B, C, and D are correct.