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(B) Let $X = (x, y, z)$. The condition for $S$ is $(x-1)^2 + (y-2)^2 + (z-3)^2 - ((x-4)^2 + (y-2)^2 + (z-7)^2) = 50$.Expanding this,we get $(x^2 - 2x

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(B) Let $X = (x, y, z)$. The condition for $S$ is $(x-1)^2 + (y-2)^2 + (z-3)^2 - ((x-4)^2 + (y-2)^2 + (z-7)^2) = 50$.Expanding this,we get $(x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9) - (x^2 - 8x + 16 + y^2 - 4y + 4 + z^2 - 14z + 49) = 50$.Simplifying,$(6x + 8z - 50) = 50$,so $6x + 8z = 100$,or $3x + 4z = 50$. This is a plane.Similarly,for $T$,we get $(x-4)^2 + (y-2)^2 + (z-7)^2 - ((x-1)^2 + (y-2)^2 + (z-3)^2) = 50$.This simplifies to $-(6x + 8z - 50) = 50$,so $6x + 8z = 0$,or $3x + 4z = 0$. This is a parallel plane.$(A)$ Since $S$ is a plane,we can always find three non-collinear points in it to form a triangle of any area,including $1$. Thus,$(A)$ is $TRUE$.$(B)$ Since $T$ is a plane,any line segment connecting two points $L, M \in T$ lies entirely within the plane $T$. Thus,$(B)$ is $TRUE$.$(C)$ The distance between the parallel planes $3x + 4z - 50 = 0$ and $3x + 4z = 0$ is $d = \frac{|50 - 0|}{\sqrt{3^2 + 4^2}} = \frac{50}{5} = 10$.For a rectangle with two vertices in $S$ and two in $T$,let the side length in the plane be $a$ and the side length connecting the planes be $b$. The distance between planes is $10$,so $b = 10$. Perimeter $2(a + 10) = 48 \implies a + 10 = 24 \implies a = 14$. Since we can choose any line of length $14$ in $S$,there are infinitely many such rectangles. Thus,$(C)$ is $TRUE$.$(D)$ For a square,$a = b = 10$. Perimeter $4a = 40 
eq 48$. However,the question asks if there is a square of perimeter $48$,which would require $a = 12$. Since the distance between planes is $10$,we can form a rectangle with sides $12$ and $10$. If we need a square,we need $a = 10$,so perimeter $40$. The statement $(D)$ claims a square of perimeter $48$ exists,which is impossible as $a$ must be $10$. Wait,checking the logic: if $a=12$ and $b=10$,it is a rectangle,not a square. Thus $(D)$ is $FALSE$.

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