JEE Advanced 2024 — Paper 1 — Question 7

Let $P=(1,2,3)$ and $Q=(4,2,7)$. For $S$, condition: $[d(X,P)]^2 - [d(X,Q)]^2 = 50$. Using distance formula: $(x-1)^2+(y-2)^2+(z-3)^2 - [(x-4)^2+(y-2)^2+(z-7)^2] = 50$. Simplify: $(x^2-2x+1)+(y^2-4y+4)+(z^2-6z+9) - [(x^2-8x+16)+(y^2-4y+4)+(z^2-14z+49)] = 50$. Cancel $x^2,y^2,z^2$: $-2x+1-4y+4-6z+9 +8x-16+4y-4+14z-49 = 50$. Combine: $(6x)+(0y)+(8z)+(1+4+9-16-4-49) = 50$ → $6x+8z-55 = 50$ → $6x+8z = 105$. So $S$ is the plane $6x+8z=105$. For $T$, condition: $[d(Y,Q)]^2 - [d(Y,P)]^2 = 50$. Similar simplification gives: $(x-4)^2+(y-2)^2+(z-7)^2 - [(x-1)^2+(y-2)^2+(z-3)^2] = 50$. Simplify: $(x^2-8x+16)+(y^2-4y+4)+(z^2-14z+49) - [(x^2-2x+1)+(y^2-4y+4)+(z^2-6z+9)] = 50$. Cancel: $-8x+16-4y+4-14z+49 +2x-1+4y-4+6z-9 = 50$. Combine: $(-6x)+(0y)+(-8z)+(16+4+49-1-4-9) = 50$ → $-6x-8z+55 = 50$ → $-6x-8z = -5$ → $6x+8z = 5$. So $T$ is the plane $6x+8z=5$. Distance between parallel planes $6x+8z=105$ and $6x+8z=5$: $d = \frac{|105-5|}{\sqrt{6^2+8^2}} = \frac{100}{10} = 10$. Option A: Choose three non-collinear points in $S$ (e.g., $(0,0,\frac{105}{8})$, $(1,0,\frac{99}{8})$, $(0,1,\frac{105}{8})$). The area of triangle formed by them can be made 1 (by appropriate scaling/choice). So A is true. Option B: $T$ is a plane, so any line segment lying in $T$ has all its points in $T$. Hence there exist two distinct points $L,M$ in $T$ such that the segment $LM$ is contained in $T$. So B is true. Option C: $S$ and $T$ are parallel planes 10 units apart. A rectangle of perimeter 48 has sides $a$ and $b$ with $2(a+b)=48$ → $a+b=24$. Choose two vertices from $S$ and two from $T$ such that the rectangle's sides are perpendicular to the planes (height = 10) and the other side length $b$ satisfies $a+b=24$. Since $a=10$, we get $b=14$.

Such a rectangle exists (e.g., take a segment of length 14 in $S$, project to $T$ to get the other two vertices). Infinitely many such rectangles exist by varying orientation. So C is true. Option D: For a square of perimeter 48, side = 12. If two vertices are in $S$ and two in $T$, the distance between the planes is 10, so the side perpendicular to the planes would be 10, but the other side must also be 12, giving a rectangle, not a square (since 10 ≠ 12). Hence no such square exists. So D is false. Thus the correct options are A, B, C.