JEE Main 2025 — 28 January, Morning Shift — Question 6

Let A be $(x, y, 0)$ in the xy-plane.

Since A is equidistant from $P(0,3,2), Q(2,0,3), R(0,0,1):$ $AP^2 = x^2 + (y-3)^2 + 4, \quad AQ^2 = (x-2)^2 + y^2 + 9, \quad AR^2 = x^2 + y^2 + 1.$

Equating $AP^2 = AR^2$ gives: $x^2 + (y-3)^2 + 4 = x^2 + y^2 + 1 \Rightarrow -6y + 13 = 1 \Rightarrow y = 2$.

Equating $AP^2 = AQ^2$ with $y=2$: $x^2 + 1 + 4 = (x-2)^2 + 4 + 9$

$\Rightarrow x^2 + 5 = x^2 -4x + 17$

$\Rightarrow -4x = -12 \Rightarrow x = 3$.

Thus $A = (3, 2, 0)$.

Compute side lengths: $AB = \sqrt{(3-1)^2 + (2-4)^2 + (0+1)^2} = 3$,

$AC = \sqrt{(3-2)^2 + (2-0)^2 + (0+2)^2} = 3$,

$BC = \sqrt{(1-2)^2 + (4-0)^2 + (-1+2)^2} = \sqrt{18} = 3\sqrt{2}$.

Since $AB = AC$, triangle is isosceles.

Also $AB^2 + AC^2 = 9+9 = 18 = BC^2$, so it is right-angled at A.

Hence Statement I is true.

Area of right triangle $ABC = \frac{1}{2} \times AB \times AC = \frac{1}{2} \times 3 \times 3 = \frac{9}{2}$.

Statement II claims area is $9\sqrt{22}$, which is false.

Therefore only Statement I is correct.