Question 7

Let $X = (x, y, z)$. The condition for $S$ is $(x-1)^2 + (y-2)^2 + (z-3)^2 - [(x-4)^2 + (y-2)^2 + (z-7)^2] = 50$, which simplifies to $6x + 8z - 55 = 50 \implies 6x + 8z = 105$. This is a plane. Similarly, the condition for $T$ is $-(6x + 8z - 55) = 50 \implies 6x + 8z = 5$. This is a parallel plane. The distance between planes $S$ and $T$ is $d = \frac{|105 - 5|}{\sqrt{6^2 + 8^2}} = \frac{100}{10} = 10$.

(A) Since $S$ is a plane, we can choose any three non-collinear points in it to form a triangle of area 1. Correct. (B) Since $T$ is a plane, any line segment connecting two points in $T$ lies entirely within $T$. Correct. (C) Let a rectangle have side lengths $a$ and $b$. If two vertices are in $S$ and two in $T$, one side (say $a$) connects the two planes. The minimum length of such a side is the distance between planes, $d = 10$. For perimeter 48, $2(a + b) = 48 \implies a + b = 24$. Since $a \ge 10$, we can choose any $a \in [10, 24)$ and find a corresponding $b > 0$. Infinitely many such rectangles exist by rotating and translating them. Correct. (D) For a square of perimeter 48, the side length is $s = 48/4 = 12$. Since the distance between the planes is $10 < 12$, we can fit a segment of length 12 between the planes to serve as a side of the square. Correct.